118478
Let \(t_n, n=1,2,3, \ldots b\) the \(n^{\text {th }}\) term of the A.P. \(5,8,11, \ldots\) Then the value of \(n\) for which \(t_n=\) 305 is
1 101
2 100
3 103
4 99
5 95
Explanation:
A Given, \(\mathrm{t}_{\mathrm{n}}=305 \text {, }\) Series in \(\mathrm{AP}=5,8,11 \ldots\). \(\mathrm{a}=5 \quad \mathrm{~d}=8-5=3\) \(\therefore \quad \mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(305=5+(\mathrm{n}-1) 3\) \(305=5+3 n-3\) \(2+3 n=305\) \(3 \mathrm{n}=303\) \(\mathrm{n}=101\)
Kerala CEE-2021
Sequence and Series
118525
The third term of a G.P. is 15 . Then the product of its first five terms is
1 \((15)^2\)
2 \((15)^3\)
3 \((15)^4\)
4 \((15)^5\)
Explanation:
D Let, a be the first term and \(\mathrm{r}\) be the common ratio of G.P. Given that, \(\mathrm{T}_3=\mathrm{ar}^2=15\) \(\therefore\) The product of its first five terms is - \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\) \(=a^5 r^{10}\) \(=\left(a r^2\right)^5\) \(=(15)^5\)
SRM JEEE-2015
Sequence and Series
118526
If the fifth term of a G.P. is 2 . Then the product of its first 9 terms is
1 256
2 512
3 1024
4 None of these
Explanation:
B Given, \(5^{\text {th }}\) term of a G.P is \(=2\) \(\mathrm{T}_5=\mathrm{ar}^4=2\) \(\therefore\) The product of first 9 terms \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\left(a r^5\right)\left(a r^6\right)\left(a r^7\right)\left(a r^8\right)\) \(=\mathrm{a}^9 \mathrm{r}^{36}\) \(=\left(\mathrm{ar}^4\right)^9\) \(=(2)^9\) \(=512\)
SRM JEEE-2016
Sequence and Series
118479
If the first term of a G.P. is 1 and the sum of \(3^{\text {rd }}\) and \(5^{\text {th }}\) terms is 90 , then the positive common ratio of the G. P. is
1 1
2 2
3 3
4 4
5 5
Explanation:
C Given, \(\mathrm{T}_1=\mathrm{a}=1\) \(\mathrm{~T}_3+\mathrm{T}_5=90\) \(\mathrm{~T}_3=\mathrm{ar}^2, \mathrm{~T}_5=\mathrm{ar}^4\) Let ' \(\mathrm{r}\) ' be the common ratio. Here, \(\quad \mathrm{T}_1=\mathrm{a}=1\) and \(\mathrm{T}_3+\mathrm{T}_5=90\) \(\therefore \quad \mathrm{ar}^2+\mathrm{ar}^4=90\) \(\mathrm{a}\left(\mathrm{r}^2+\mathrm{r}^4\right)=90 \quad[\because \mathrm{a}=1]\) \(\therefore \quad \mathrm{r}^4+\mathrm{r}^2-90=0\) Which is a quadratic equation in \(\mathrm{r}^2\). \(\therefore \quad \mathrm{r}^2=\frac{-1 \pm \sqrt{(1)^2-4 \times 1 \times(-90)}}{2 \times 1}\) \(\mathrm{r}^2=\frac{-1 \pm \sqrt{1+360}}{2}=\frac{-1 \pm \sqrt{361}}{2}\) \(\mathrm{r}^2=\frac{-1 \pm 19}{2}\) Either, \(\mathrm{r}^2=\frac{-1+19}{2}\) i.e, \(\mathrm{r}^2=\frac{18}{2}\) \(\Rightarrow \quad r^2=9 \Rightarrow r \pm 3\) Or \(\mathrm{r}^2=\frac{-1-19}{2} \text { i.e, } \mathrm{r}^2=\frac{-20}{2}\) \(\mathrm{r}^2=-10 \quad(\because \text { Which is not possible })\)Thus, common ratio of G.P. is \(\pm 3\).
Kerala CEE-2021
Sequence and Series
118480
In an A.P. the difference between the last and the first terms is 632 and the common difference is 4 . Then the number of terms in the A. P. is
1 157
2 160
3 158
4 159
5 140
Explanation:
D Given, Last term \(=\mathrm{T}_{\mathrm{n}}\) First term \(=\mathrm{a}\) \(\mathrm{T}_{\mathrm{n}}-\mathrm{a}=632, \mathrm{~d}=4, \mathrm{n}=\) ? \(n^{\text {th }}\) terms, \(T_n=a+(n-1) d\) Now, \(\quad a+(n-1) d-a=632\) \((n-1) 4=632\) \(\mathrm{n}-1=\frac{632}{4}=158\) \(\mathrm{n}-1=158\) \(\Rightarrow \quad \mathrm{n}=159\)
118478
Let \(t_n, n=1,2,3, \ldots b\) the \(n^{\text {th }}\) term of the A.P. \(5,8,11, \ldots\) Then the value of \(n\) for which \(t_n=\) 305 is
1 101
2 100
3 103
4 99
5 95
Explanation:
A Given, \(\mathrm{t}_{\mathrm{n}}=305 \text {, }\) Series in \(\mathrm{AP}=5,8,11 \ldots\). \(\mathrm{a}=5 \quad \mathrm{~d}=8-5=3\) \(\therefore \quad \mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(305=5+(\mathrm{n}-1) 3\) \(305=5+3 n-3\) \(2+3 n=305\) \(3 \mathrm{n}=303\) \(\mathrm{n}=101\)
Kerala CEE-2021
Sequence and Series
118525
The third term of a G.P. is 15 . Then the product of its first five terms is
1 \((15)^2\)
2 \((15)^3\)
3 \((15)^4\)
4 \((15)^5\)
Explanation:
D Let, a be the first term and \(\mathrm{r}\) be the common ratio of G.P. Given that, \(\mathrm{T}_3=\mathrm{ar}^2=15\) \(\therefore\) The product of its first five terms is - \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\) \(=a^5 r^{10}\) \(=\left(a r^2\right)^5\) \(=(15)^5\)
SRM JEEE-2015
Sequence and Series
118526
If the fifth term of a G.P. is 2 . Then the product of its first 9 terms is
1 256
2 512
3 1024
4 None of these
Explanation:
B Given, \(5^{\text {th }}\) term of a G.P is \(=2\) \(\mathrm{T}_5=\mathrm{ar}^4=2\) \(\therefore\) The product of first 9 terms \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\left(a r^5\right)\left(a r^6\right)\left(a r^7\right)\left(a r^8\right)\) \(=\mathrm{a}^9 \mathrm{r}^{36}\) \(=\left(\mathrm{ar}^4\right)^9\) \(=(2)^9\) \(=512\)
SRM JEEE-2016
Sequence and Series
118479
If the first term of a G.P. is 1 and the sum of \(3^{\text {rd }}\) and \(5^{\text {th }}\) terms is 90 , then the positive common ratio of the G. P. is
1 1
2 2
3 3
4 4
5 5
Explanation:
C Given, \(\mathrm{T}_1=\mathrm{a}=1\) \(\mathrm{~T}_3+\mathrm{T}_5=90\) \(\mathrm{~T}_3=\mathrm{ar}^2, \mathrm{~T}_5=\mathrm{ar}^4\) Let ' \(\mathrm{r}\) ' be the common ratio. Here, \(\quad \mathrm{T}_1=\mathrm{a}=1\) and \(\mathrm{T}_3+\mathrm{T}_5=90\) \(\therefore \quad \mathrm{ar}^2+\mathrm{ar}^4=90\) \(\mathrm{a}\left(\mathrm{r}^2+\mathrm{r}^4\right)=90 \quad[\because \mathrm{a}=1]\) \(\therefore \quad \mathrm{r}^4+\mathrm{r}^2-90=0\) Which is a quadratic equation in \(\mathrm{r}^2\). \(\therefore \quad \mathrm{r}^2=\frac{-1 \pm \sqrt{(1)^2-4 \times 1 \times(-90)}}{2 \times 1}\) \(\mathrm{r}^2=\frac{-1 \pm \sqrt{1+360}}{2}=\frac{-1 \pm \sqrt{361}}{2}\) \(\mathrm{r}^2=\frac{-1 \pm 19}{2}\) Either, \(\mathrm{r}^2=\frac{-1+19}{2}\) i.e, \(\mathrm{r}^2=\frac{18}{2}\) \(\Rightarrow \quad r^2=9 \Rightarrow r \pm 3\) Or \(\mathrm{r}^2=\frac{-1-19}{2} \text { i.e, } \mathrm{r}^2=\frac{-20}{2}\) \(\mathrm{r}^2=-10 \quad(\because \text { Which is not possible })\)Thus, common ratio of G.P. is \(\pm 3\).
Kerala CEE-2021
Sequence and Series
118480
In an A.P. the difference between the last and the first terms is 632 and the common difference is 4 . Then the number of terms in the A. P. is
1 157
2 160
3 158
4 159
5 140
Explanation:
D Given, Last term \(=\mathrm{T}_{\mathrm{n}}\) First term \(=\mathrm{a}\) \(\mathrm{T}_{\mathrm{n}}-\mathrm{a}=632, \mathrm{~d}=4, \mathrm{n}=\) ? \(n^{\text {th }}\) terms, \(T_n=a+(n-1) d\) Now, \(\quad a+(n-1) d-a=632\) \((n-1) 4=632\) \(\mathrm{n}-1=\frac{632}{4}=158\) \(\mathrm{n}-1=158\) \(\Rightarrow \quad \mathrm{n}=159\)
118478
Let \(t_n, n=1,2,3, \ldots b\) the \(n^{\text {th }}\) term of the A.P. \(5,8,11, \ldots\) Then the value of \(n\) for which \(t_n=\) 305 is
1 101
2 100
3 103
4 99
5 95
Explanation:
A Given, \(\mathrm{t}_{\mathrm{n}}=305 \text {, }\) Series in \(\mathrm{AP}=5,8,11 \ldots\). \(\mathrm{a}=5 \quad \mathrm{~d}=8-5=3\) \(\therefore \quad \mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(305=5+(\mathrm{n}-1) 3\) \(305=5+3 n-3\) \(2+3 n=305\) \(3 \mathrm{n}=303\) \(\mathrm{n}=101\)
Kerala CEE-2021
Sequence and Series
118525
The third term of a G.P. is 15 . Then the product of its first five terms is
1 \((15)^2\)
2 \((15)^3\)
3 \((15)^4\)
4 \((15)^5\)
Explanation:
D Let, a be the first term and \(\mathrm{r}\) be the common ratio of G.P. Given that, \(\mathrm{T}_3=\mathrm{ar}^2=15\) \(\therefore\) The product of its first five terms is - \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\) \(=a^5 r^{10}\) \(=\left(a r^2\right)^5\) \(=(15)^5\)
SRM JEEE-2015
Sequence and Series
118526
If the fifth term of a G.P. is 2 . Then the product of its first 9 terms is
1 256
2 512
3 1024
4 None of these
Explanation:
B Given, \(5^{\text {th }}\) term of a G.P is \(=2\) \(\mathrm{T}_5=\mathrm{ar}^4=2\) \(\therefore\) The product of first 9 terms \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\left(a r^5\right)\left(a r^6\right)\left(a r^7\right)\left(a r^8\right)\) \(=\mathrm{a}^9 \mathrm{r}^{36}\) \(=\left(\mathrm{ar}^4\right)^9\) \(=(2)^9\) \(=512\)
SRM JEEE-2016
Sequence and Series
118479
If the first term of a G.P. is 1 and the sum of \(3^{\text {rd }}\) and \(5^{\text {th }}\) terms is 90 , then the positive common ratio of the G. P. is
1 1
2 2
3 3
4 4
5 5
Explanation:
C Given, \(\mathrm{T}_1=\mathrm{a}=1\) \(\mathrm{~T}_3+\mathrm{T}_5=90\) \(\mathrm{~T}_3=\mathrm{ar}^2, \mathrm{~T}_5=\mathrm{ar}^4\) Let ' \(\mathrm{r}\) ' be the common ratio. Here, \(\quad \mathrm{T}_1=\mathrm{a}=1\) and \(\mathrm{T}_3+\mathrm{T}_5=90\) \(\therefore \quad \mathrm{ar}^2+\mathrm{ar}^4=90\) \(\mathrm{a}\left(\mathrm{r}^2+\mathrm{r}^4\right)=90 \quad[\because \mathrm{a}=1]\) \(\therefore \quad \mathrm{r}^4+\mathrm{r}^2-90=0\) Which is a quadratic equation in \(\mathrm{r}^2\). \(\therefore \quad \mathrm{r}^2=\frac{-1 \pm \sqrt{(1)^2-4 \times 1 \times(-90)}}{2 \times 1}\) \(\mathrm{r}^2=\frac{-1 \pm \sqrt{1+360}}{2}=\frac{-1 \pm \sqrt{361}}{2}\) \(\mathrm{r}^2=\frac{-1 \pm 19}{2}\) Either, \(\mathrm{r}^2=\frac{-1+19}{2}\) i.e, \(\mathrm{r}^2=\frac{18}{2}\) \(\Rightarrow \quad r^2=9 \Rightarrow r \pm 3\) Or \(\mathrm{r}^2=\frac{-1-19}{2} \text { i.e, } \mathrm{r}^2=\frac{-20}{2}\) \(\mathrm{r}^2=-10 \quad(\because \text { Which is not possible })\)Thus, common ratio of G.P. is \(\pm 3\).
Kerala CEE-2021
Sequence and Series
118480
In an A.P. the difference between the last and the first terms is 632 and the common difference is 4 . Then the number of terms in the A. P. is
1 157
2 160
3 158
4 159
5 140
Explanation:
D Given, Last term \(=\mathrm{T}_{\mathrm{n}}\) First term \(=\mathrm{a}\) \(\mathrm{T}_{\mathrm{n}}-\mathrm{a}=632, \mathrm{~d}=4, \mathrm{n}=\) ? \(n^{\text {th }}\) terms, \(T_n=a+(n-1) d\) Now, \(\quad a+(n-1) d-a=632\) \((n-1) 4=632\) \(\mathrm{n}-1=\frac{632}{4}=158\) \(\mathrm{n}-1=158\) \(\Rightarrow \quad \mathrm{n}=159\)
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Sequence and Series
118478
Let \(t_n, n=1,2,3, \ldots b\) the \(n^{\text {th }}\) term of the A.P. \(5,8,11, \ldots\) Then the value of \(n\) for which \(t_n=\) 305 is
1 101
2 100
3 103
4 99
5 95
Explanation:
A Given, \(\mathrm{t}_{\mathrm{n}}=305 \text {, }\) Series in \(\mathrm{AP}=5,8,11 \ldots\). \(\mathrm{a}=5 \quad \mathrm{~d}=8-5=3\) \(\therefore \quad \mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(305=5+(\mathrm{n}-1) 3\) \(305=5+3 n-3\) \(2+3 n=305\) \(3 \mathrm{n}=303\) \(\mathrm{n}=101\)
Kerala CEE-2021
Sequence and Series
118525
The third term of a G.P. is 15 . Then the product of its first five terms is
1 \((15)^2\)
2 \((15)^3\)
3 \((15)^4\)
4 \((15)^5\)
Explanation:
D Let, a be the first term and \(\mathrm{r}\) be the common ratio of G.P. Given that, \(\mathrm{T}_3=\mathrm{ar}^2=15\) \(\therefore\) The product of its first five terms is - \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\) \(=a^5 r^{10}\) \(=\left(a r^2\right)^5\) \(=(15)^5\)
SRM JEEE-2015
Sequence and Series
118526
If the fifth term of a G.P. is 2 . Then the product of its first 9 terms is
1 256
2 512
3 1024
4 None of these
Explanation:
B Given, \(5^{\text {th }}\) term of a G.P is \(=2\) \(\mathrm{T}_5=\mathrm{ar}^4=2\) \(\therefore\) The product of first 9 terms \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\left(a r^5\right)\left(a r^6\right)\left(a r^7\right)\left(a r^8\right)\) \(=\mathrm{a}^9 \mathrm{r}^{36}\) \(=\left(\mathrm{ar}^4\right)^9\) \(=(2)^9\) \(=512\)
SRM JEEE-2016
Sequence and Series
118479
If the first term of a G.P. is 1 and the sum of \(3^{\text {rd }}\) and \(5^{\text {th }}\) terms is 90 , then the positive common ratio of the G. P. is
1 1
2 2
3 3
4 4
5 5
Explanation:
C Given, \(\mathrm{T}_1=\mathrm{a}=1\) \(\mathrm{~T}_3+\mathrm{T}_5=90\) \(\mathrm{~T}_3=\mathrm{ar}^2, \mathrm{~T}_5=\mathrm{ar}^4\) Let ' \(\mathrm{r}\) ' be the common ratio. Here, \(\quad \mathrm{T}_1=\mathrm{a}=1\) and \(\mathrm{T}_3+\mathrm{T}_5=90\) \(\therefore \quad \mathrm{ar}^2+\mathrm{ar}^4=90\) \(\mathrm{a}\left(\mathrm{r}^2+\mathrm{r}^4\right)=90 \quad[\because \mathrm{a}=1]\) \(\therefore \quad \mathrm{r}^4+\mathrm{r}^2-90=0\) Which is a quadratic equation in \(\mathrm{r}^2\). \(\therefore \quad \mathrm{r}^2=\frac{-1 \pm \sqrt{(1)^2-4 \times 1 \times(-90)}}{2 \times 1}\) \(\mathrm{r}^2=\frac{-1 \pm \sqrt{1+360}}{2}=\frac{-1 \pm \sqrt{361}}{2}\) \(\mathrm{r}^2=\frac{-1 \pm 19}{2}\) Either, \(\mathrm{r}^2=\frac{-1+19}{2}\) i.e, \(\mathrm{r}^2=\frac{18}{2}\) \(\Rightarrow \quad r^2=9 \Rightarrow r \pm 3\) Or \(\mathrm{r}^2=\frac{-1-19}{2} \text { i.e, } \mathrm{r}^2=\frac{-20}{2}\) \(\mathrm{r}^2=-10 \quad(\because \text { Which is not possible })\)Thus, common ratio of G.P. is \(\pm 3\).
Kerala CEE-2021
Sequence and Series
118480
In an A.P. the difference between the last and the first terms is 632 and the common difference is 4 . Then the number of terms in the A. P. is
1 157
2 160
3 158
4 159
5 140
Explanation:
D Given, Last term \(=\mathrm{T}_{\mathrm{n}}\) First term \(=\mathrm{a}\) \(\mathrm{T}_{\mathrm{n}}-\mathrm{a}=632, \mathrm{~d}=4, \mathrm{n}=\) ? \(n^{\text {th }}\) terms, \(T_n=a+(n-1) d\) Now, \(\quad a+(n-1) d-a=632\) \((n-1) 4=632\) \(\mathrm{n}-1=\frac{632}{4}=158\) \(\mathrm{n}-1=158\) \(\Rightarrow \quad \mathrm{n}=159\)
118478
Let \(t_n, n=1,2,3, \ldots b\) the \(n^{\text {th }}\) term of the A.P. \(5,8,11, \ldots\) Then the value of \(n\) for which \(t_n=\) 305 is
1 101
2 100
3 103
4 99
5 95
Explanation:
A Given, \(\mathrm{t}_{\mathrm{n}}=305 \text {, }\) Series in \(\mathrm{AP}=5,8,11 \ldots\). \(\mathrm{a}=5 \quad \mathrm{~d}=8-5=3\) \(\therefore \quad \mathrm{t}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(305=5+(\mathrm{n}-1) 3\) \(305=5+3 n-3\) \(2+3 n=305\) \(3 \mathrm{n}=303\) \(\mathrm{n}=101\)
Kerala CEE-2021
Sequence and Series
118525
The third term of a G.P. is 15 . Then the product of its first five terms is
1 \((15)^2\)
2 \((15)^3\)
3 \((15)^4\)
4 \((15)^5\)
Explanation:
D Let, a be the first term and \(\mathrm{r}\) be the common ratio of G.P. Given that, \(\mathrm{T}_3=\mathrm{ar}^2=15\) \(\therefore\) The product of its first five terms is - \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\) \(=a^5 r^{10}\) \(=\left(a r^2\right)^5\) \(=(15)^5\)
SRM JEEE-2015
Sequence and Series
118526
If the fifth term of a G.P. is 2 . Then the product of its first 9 terms is
1 256
2 512
3 1024
4 None of these
Explanation:
B Given, \(5^{\text {th }}\) term of a G.P is \(=2\) \(\mathrm{T}_5=\mathrm{ar}^4=2\) \(\therefore\) The product of first 9 terms \(=(a)(a r)\left(a r^2\right)\left(a r^3\right)\left(a r^4\right)\left(a r^5\right)\left(a r^6\right)\left(a r^7\right)\left(a r^8\right)\) \(=\mathrm{a}^9 \mathrm{r}^{36}\) \(=\left(\mathrm{ar}^4\right)^9\) \(=(2)^9\) \(=512\)
SRM JEEE-2016
Sequence and Series
118479
If the first term of a G.P. is 1 and the sum of \(3^{\text {rd }}\) and \(5^{\text {th }}\) terms is 90 , then the positive common ratio of the G. P. is
1 1
2 2
3 3
4 4
5 5
Explanation:
C Given, \(\mathrm{T}_1=\mathrm{a}=1\) \(\mathrm{~T}_3+\mathrm{T}_5=90\) \(\mathrm{~T}_3=\mathrm{ar}^2, \mathrm{~T}_5=\mathrm{ar}^4\) Let ' \(\mathrm{r}\) ' be the common ratio. Here, \(\quad \mathrm{T}_1=\mathrm{a}=1\) and \(\mathrm{T}_3+\mathrm{T}_5=90\) \(\therefore \quad \mathrm{ar}^2+\mathrm{ar}^4=90\) \(\mathrm{a}\left(\mathrm{r}^2+\mathrm{r}^4\right)=90 \quad[\because \mathrm{a}=1]\) \(\therefore \quad \mathrm{r}^4+\mathrm{r}^2-90=0\) Which is a quadratic equation in \(\mathrm{r}^2\). \(\therefore \quad \mathrm{r}^2=\frac{-1 \pm \sqrt{(1)^2-4 \times 1 \times(-90)}}{2 \times 1}\) \(\mathrm{r}^2=\frac{-1 \pm \sqrt{1+360}}{2}=\frac{-1 \pm \sqrt{361}}{2}\) \(\mathrm{r}^2=\frac{-1 \pm 19}{2}\) Either, \(\mathrm{r}^2=\frac{-1+19}{2}\) i.e, \(\mathrm{r}^2=\frac{18}{2}\) \(\Rightarrow \quad r^2=9 \Rightarrow r \pm 3\) Or \(\mathrm{r}^2=\frac{-1-19}{2} \text { i.e, } \mathrm{r}^2=\frac{-20}{2}\) \(\mathrm{r}^2=-10 \quad(\because \text { Which is not possible })\)Thus, common ratio of G.P. is \(\pm 3\).
Kerala CEE-2021
Sequence and Series
118480
In an A.P. the difference between the last and the first terms is 632 and the common difference is 4 . Then the number of terms in the A. P. is
1 157
2 160
3 158
4 159
5 140
Explanation:
D Given, Last term \(=\mathrm{T}_{\mathrm{n}}\) First term \(=\mathrm{a}\) \(\mathrm{T}_{\mathrm{n}}-\mathrm{a}=632, \mathrm{~d}=4, \mathrm{n}=\) ? \(n^{\text {th }}\) terms, \(T_n=a+(n-1) d\) Now, \(\quad a+(n-1) d-a=632\) \((n-1) 4=632\) \(\mathrm{n}-1=\frac{632}{4}=158\) \(\mathrm{n}-1=158\) \(\Rightarrow \quad \mathrm{n}=159\)