Arithmetic Progression
Sequence and Series

118489 The \(30^{\text {th }}\) term to the arithmetic progression 10 , 7,4 is

1 -97
2 -87
3 -77
4 -67
5 -57
Sequence and Series

118535 For a GP, \(a_n=3\left(2^n\right), \forall n \in N\). Find the common ratio.

1 2
2 \(1 / 2\)
3 3
4 \(1 / 3\)
Sequence and Series

118490 Three number \(x, y\) and \(z\) are in arithmetic progression. If \(x+y+z=-3\) and \(x y z=8\), then \(\mathbf{x}^2+\mathrm{y}^2+\mathrm{z}^2\) is equal to

1 9
2 10
3 21
4 20
5 1
Sequence and Series

118491 In an \(A P\), the \(6^{\text {th }}\) term is 52 and the \(1^{\text {th }}\) term is 112. Then, the common difference is equal to

1 4
2 20
3 12
4 8
5 6
Sequence and Series

118492 If \(a_1, a_2, a_3, \ldots, a_{20}\) are in AP and \(a_1+a_{20}=45\), then \(\mathbf{a}_1+\mathbf{a}_2+\mathbf{a}_3+\ldots+\mathbf{a}_{20}\) is equal to

1 90
2 900
3 350
4 450
5 730
Sequence and Series

118489 The \(30^{\text {th }}\) term to the arithmetic progression 10 , 7,4 is

1 -97
2 -87
3 -77
4 -67
5 -57
Sequence and Series

118535 For a GP, \(a_n=3\left(2^n\right), \forall n \in N\). Find the common ratio.

1 2
2 \(1 / 2\)
3 3
4 \(1 / 3\)
Sequence and Series

118490 Three number \(x, y\) and \(z\) are in arithmetic progression. If \(x+y+z=-3\) and \(x y z=8\), then \(\mathbf{x}^2+\mathrm{y}^2+\mathrm{z}^2\) is equal to

1 9
2 10
3 21
4 20
5 1
Sequence and Series

118491 In an \(A P\), the \(6^{\text {th }}\) term is 52 and the \(1^{\text {th }}\) term is 112. Then, the common difference is equal to

1 4
2 20
3 12
4 8
5 6
Sequence and Series

118492 If \(a_1, a_2, a_3, \ldots, a_{20}\) are in AP and \(a_1+a_{20}=45\), then \(\mathbf{a}_1+\mathbf{a}_2+\mathbf{a}_3+\ldots+\mathbf{a}_{20}\) is equal to

1 90
2 900
3 350
4 450
5 730
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Sequence and Series

118489 The \(30^{\text {th }}\) term to the arithmetic progression 10 , 7,4 is

1 -97
2 -87
3 -77
4 -67
5 -57
Sequence and Series

118535 For a GP, \(a_n=3\left(2^n\right), \forall n \in N\). Find the common ratio.

1 2
2 \(1 / 2\)
3 3
4 \(1 / 3\)
Sequence and Series

118490 Three number \(x, y\) and \(z\) are in arithmetic progression. If \(x+y+z=-3\) and \(x y z=8\), then \(\mathbf{x}^2+\mathrm{y}^2+\mathrm{z}^2\) is equal to

1 9
2 10
3 21
4 20
5 1
Sequence and Series

118491 In an \(A P\), the \(6^{\text {th }}\) term is 52 and the \(1^{\text {th }}\) term is 112. Then, the common difference is equal to

1 4
2 20
3 12
4 8
5 6
Sequence and Series

118492 If \(a_1, a_2, a_3, \ldots, a_{20}\) are in AP and \(a_1+a_{20}=45\), then \(\mathbf{a}_1+\mathbf{a}_2+\mathbf{a}_3+\ldots+\mathbf{a}_{20}\) is equal to

1 90
2 900
3 350
4 450
5 730
Sequence and Series

118489 The \(30^{\text {th }}\) term to the arithmetic progression 10 , 7,4 is

1 -97
2 -87
3 -77
4 -67
5 -57
Sequence and Series

118535 For a GP, \(a_n=3\left(2^n\right), \forall n \in N\). Find the common ratio.

1 2
2 \(1 / 2\)
3 3
4 \(1 / 3\)
Sequence and Series

118490 Three number \(x, y\) and \(z\) are in arithmetic progression. If \(x+y+z=-3\) and \(x y z=8\), then \(\mathbf{x}^2+\mathrm{y}^2+\mathrm{z}^2\) is equal to

1 9
2 10
3 21
4 20
5 1
Sequence and Series

118491 In an \(A P\), the \(6^{\text {th }}\) term is 52 and the \(1^{\text {th }}\) term is 112. Then, the common difference is equal to

1 4
2 20
3 12
4 8
5 6
Sequence and Series

118492 If \(a_1, a_2, a_3, \ldots, a_{20}\) are in AP and \(a_1+a_{20}=45\), then \(\mathbf{a}_1+\mathbf{a}_2+\mathbf{a}_3+\ldots+\mathbf{a}_{20}\) is equal to

1 90
2 900
3 350
4 450
5 730
Sequence and Series

118489 The \(30^{\text {th }}\) term to the arithmetic progression 10 , 7,4 is

1 -97
2 -87
3 -77
4 -67
5 -57
Sequence and Series

118535 For a GP, \(a_n=3\left(2^n\right), \forall n \in N\). Find the common ratio.

1 2
2 \(1 / 2\)
3 3
4 \(1 / 3\)
Sequence and Series

118490 Three number \(x, y\) and \(z\) are in arithmetic progression. If \(x+y+z=-3\) and \(x y z=8\), then \(\mathbf{x}^2+\mathrm{y}^2+\mathrm{z}^2\) is equal to

1 9
2 10
3 21
4 20
5 1
Sequence and Series

118491 In an \(A P\), the \(6^{\text {th }}\) term is 52 and the \(1^{\text {th }}\) term is 112. Then, the common difference is equal to

1 4
2 20
3 12
4 8
5 6
Sequence and Series

118492 If \(a_1, a_2, a_3, \ldots, a_{20}\) are in AP and \(a_1+a_{20}=45\), then \(\mathbf{a}_1+\mathbf{a}_2+\mathbf{a}_3+\ldots+\mathbf{a}_{20}\) is equal to

1 90
2 900
3 350
4 450
5 730