118481
If the \(10^{\text {th }}\) and \(12^{\text {th }}\) terms of an A. P. are respectively 15 and 21 , then the common difference of the A.P. is
1 -6
2 4
3 6
4 -3
5 3
Explanation:
E Given, \(\mathrm{T}_{10}=15, \mathrm{~T}_{12}=21\), For series AP, \(\mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(\mathrm{T}_{10}=\mathrm{a}+9 \mathrm{~d}=15\) And, \(\quad \mathrm{T}_{12}=21\) \(\therefore \quad \mathrm{T}_{12}=\mathrm{a}+11 \mathrm{~d}=21\) Solving equation (i) and (ii), we get- \(\mathrm{d}=3\)Thus, the common difference is 3 .
Kerala CEE-2021
Sequence and Series
118482
The first term of a G.P. is 3 and the common ratio is 2 . Then the sum of first eight terms of the G.P. is
118483
If \(a_1, a_2, a_3, \ldots ., a_n\) are in A. P. with \(a_1=3, a_n=\) 39 and \(a_1+a_2+\ldots+a_n=210\), then the value of \(n\) is equal to
1 8
2 10
3 11
4 13
5 15
Explanation:
B Given, \(a_1, a_2, \ldots . . a_n \Rightarrow A P\) First term \(\left(a_1\right)=3\), Last term \(\left(a_n\right)=39\) \(a_1+a_2+\ldots \ldots+a_n=210,\) Sum of \(\mathrm{n}^{\text {th }}\) term in A.P, \(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}_1+\mathrm{a}_{\mathrm{n}}\right)\) \(\frac{\mathrm{n}}{2}(3+39)=210\) \(\frac{\mathrm{n}}{2} \times 42=210\) \(21 \mathrm{n}=210\) \(\mathrm{n}=\frac{210}{21}\) \(\mathrm{n}=10\)
Kerala CEE-2021
Sequence and Series
118484
In an A.P., the first is 3 and the last terms is 17. The sum of all the terms in the sequence is 70 . Then the number of terms in the arithmetic sequence is
1 7
2 5
3 9
4 6
5 8
Explanation:
A Given, First term \((\mathrm{a})=3\), Last term \(\left(\mathrm{a}_{\mathrm{n}}\right)=17, \mathrm{~S}_{\mathrm{n}}=70\) Sum of series AP, \(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]\) \(70=\frac{\mathrm{n}}{2}[3+17]\) \(70=\frac{\mathrm{n}}{2} \times 20\) \(\mathrm{n}=7\)
118481
If the \(10^{\text {th }}\) and \(12^{\text {th }}\) terms of an A. P. are respectively 15 and 21 , then the common difference of the A.P. is
1 -6
2 4
3 6
4 -3
5 3
Explanation:
E Given, \(\mathrm{T}_{10}=15, \mathrm{~T}_{12}=21\), For series AP, \(\mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(\mathrm{T}_{10}=\mathrm{a}+9 \mathrm{~d}=15\) And, \(\quad \mathrm{T}_{12}=21\) \(\therefore \quad \mathrm{T}_{12}=\mathrm{a}+11 \mathrm{~d}=21\) Solving equation (i) and (ii), we get- \(\mathrm{d}=3\)Thus, the common difference is 3 .
Kerala CEE-2021
Sequence and Series
118482
The first term of a G.P. is 3 and the common ratio is 2 . Then the sum of first eight terms of the G.P. is
118483
If \(a_1, a_2, a_3, \ldots ., a_n\) are in A. P. with \(a_1=3, a_n=\) 39 and \(a_1+a_2+\ldots+a_n=210\), then the value of \(n\) is equal to
1 8
2 10
3 11
4 13
5 15
Explanation:
B Given, \(a_1, a_2, \ldots . . a_n \Rightarrow A P\) First term \(\left(a_1\right)=3\), Last term \(\left(a_n\right)=39\) \(a_1+a_2+\ldots \ldots+a_n=210,\) Sum of \(\mathrm{n}^{\text {th }}\) term in A.P, \(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}_1+\mathrm{a}_{\mathrm{n}}\right)\) \(\frac{\mathrm{n}}{2}(3+39)=210\) \(\frac{\mathrm{n}}{2} \times 42=210\) \(21 \mathrm{n}=210\) \(\mathrm{n}=\frac{210}{21}\) \(\mathrm{n}=10\)
Kerala CEE-2021
Sequence and Series
118484
In an A.P., the first is 3 and the last terms is 17. The sum of all the terms in the sequence is 70 . Then the number of terms in the arithmetic sequence is
1 7
2 5
3 9
4 6
5 8
Explanation:
A Given, First term \((\mathrm{a})=3\), Last term \(\left(\mathrm{a}_{\mathrm{n}}\right)=17, \mathrm{~S}_{\mathrm{n}}=70\) Sum of series AP, \(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]\) \(70=\frac{\mathrm{n}}{2}[3+17]\) \(70=\frac{\mathrm{n}}{2} \times 20\) \(\mathrm{n}=7\)
118481
If the \(10^{\text {th }}\) and \(12^{\text {th }}\) terms of an A. P. are respectively 15 and 21 , then the common difference of the A.P. is
1 -6
2 4
3 6
4 -3
5 3
Explanation:
E Given, \(\mathrm{T}_{10}=15, \mathrm{~T}_{12}=21\), For series AP, \(\mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(\mathrm{T}_{10}=\mathrm{a}+9 \mathrm{~d}=15\) And, \(\quad \mathrm{T}_{12}=21\) \(\therefore \quad \mathrm{T}_{12}=\mathrm{a}+11 \mathrm{~d}=21\) Solving equation (i) and (ii), we get- \(\mathrm{d}=3\)Thus, the common difference is 3 .
Kerala CEE-2021
Sequence and Series
118482
The first term of a G.P. is 3 and the common ratio is 2 . Then the sum of first eight terms of the G.P. is
118483
If \(a_1, a_2, a_3, \ldots ., a_n\) are in A. P. with \(a_1=3, a_n=\) 39 and \(a_1+a_2+\ldots+a_n=210\), then the value of \(n\) is equal to
1 8
2 10
3 11
4 13
5 15
Explanation:
B Given, \(a_1, a_2, \ldots . . a_n \Rightarrow A P\) First term \(\left(a_1\right)=3\), Last term \(\left(a_n\right)=39\) \(a_1+a_2+\ldots \ldots+a_n=210,\) Sum of \(\mathrm{n}^{\text {th }}\) term in A.P, \(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}_1+\mathrm{a}_{\mathrm{n}}\right)\) \(\frac{\mathrm{n}}{2}(3+39)=210\) \(\frac{\mathrm{n}}{2} \times 42=210\) \(21 \mathrm{n}=210\) \(\mathrm{n}=\frac{210}{21}\) \(\mathrm{n}=10\)
Kerala CEE-2021
Sequence and Series
118484
In an A.P., the first is 3 and the last terms is 17. The sum of all the terms in the sequence is 70 . Then the number of terms in the arithmetic sequence is
1 7
2 5
3 9
4 6
5 8
Explanation:
A Given, First term \((\mathrm{a})=3\), Last term \(\left(\mathrm{a}_{\mathrm{n}}\right)=17, \mathrm{~S}_{\mathrm{n}}=70\) Sum of series AP, \(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]\) \(70=\frac{\mathrm{n}}{2}[3+17]\) \(70=\frac{\mathrm{n}}{2} \times 20\) \(\mathrm{n}=7\)
118481
If the \(10^{\text {th }}\) and \(12^{\text {th }}\) terms of an A. P. are respectively 15 and 21 , then the common difference of the A.P. is
1 -6
2 4
3 6
4 -3
5 3
Explanation:
E Given, \(\mathrm{T}_{10}=15, \mathrm{~T}_{12}=21\), For series AP, \(\mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}\) \(\mathrm{T}_{10}=\mathrm{a}+9 \mathrm{~d}=15\) And, \(\quad \mathrm{T}_{12}=21\) \(\therefore \quad \mathrm{T}_{12}=\mathrm{a}+11 \mathrm{~d}=21\) Solving equation (i) and (ii), we get- \(\mathrm{d}=3\)Thus, the common difference is 3 .
Kerala CEE-2021
Sequence and Series
118482
The first term of a G.P. is 3 and the common ratio is 2 . Then the sum of first eight terms of the G.P. is
118483
If \(a_1, a_2, a_3, \ldots ., a_n\) are in A. P. with \(a_1=3, a_n=\) 39 and \(a_1+a_2+\ldots+a_n=210\), then the value of \(n\) is equal to
1 8
2 10
3 11
4 13
5 15
Explanation:
B Given, \(a_1, a_2, \ldots . . a_n \Rightarrow A P\) First term \(\left(a_1\right)=3\), Last term \(\left(a_n\right)=39\) \(a_1+a_2+\ldots \ldots+a_n=210,\) Sum of \(\mathrm{n}^{\text {th }}\) term in A.P, \(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}_1+\mathrm{a}_{\mathrm{n}}\right)\) \(\frac{\mathrm{n}}{2}(3+39)=210\) \(\frac{\mathrm{n}}{2} \times 42=210\) \(21 \mathrm{n}=210\) \(\mathrm{n}=\frac{210}{21}\) \(\mathrm{n}=10\)
Kerala CEE-2021
Sequence and Series
118484
In an A.P., the first is 3 and the last terms is 17. The sum of all the terms in the sequence is 70 . Then the number of terms in the arithmetic sequence is
1 7
2 5
3 9
4 6
5 8
Explanation:
A Given, First term \((\mathrm{a})=3\), Last term \(\left(\mathrm{a}_{\mathrm{n}}\right)=17, \mathrm{~S}_{\mathrm{n}}=70\) Sum of series AP, \(\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]\) \(70=\frac{\mathrm{n}}{2}[3+17]\) \(70=\frac{\mathrm{n}}{2} \times 20\) \(\mathrm{n}=7\)
