Arithmetic Progression
Sequence and Series

118493 If \(a_1, a_2, a_3, a_4\) are in \(A P\), then \(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\frac{1}{\sqrt{a_3}+\sqrt{a_4}}\) is equal to

1 \(\frac{\sqrt{a_4}-\sqrt{a_1}}{a_3-a_2}\)
2 \(\frac{\mathrm{a}_4-\mathrm{a}_1}{\mathrm{a}_3-\mathrm{a}_2}\)
3 \(\frac{\mathrm{a}_3-\mathrm{a}_2}{\sqrt{\mathrm{a}_4}-\sqrt{\mathrm{a}_1}}\)
4 \(\frac{\mathrm{a}_1-\mathrm{a}_4}{\mathrm{a}_3-\mathrm{a}_1}\)
5 \(\frac{a_5-a_0}{a_1-a_4}\)
Sequence and Series

118494 If the product of five consecutive terms of a GP is \(\frac{243}{32}\), then the middle term is

1 \(2 / 3\)
2 \(3 / 2\)
3 \(4 / 3\)
4 \(3 / 4\)
5 1
Sequence and Series

118495 If \(6^{\text {th }}\) term of a GP is 2 , then the product of first 11 terms of the GP is equal to

1 512
2 1024
3 2048
4 256
5 32
Sequence and Series

118496 If a,b,c are in \(A P\) and if their squares taken in the same order form a GP, then \((a+c)^4\) is equal to.

1 \(16 \mathrm{a}^2 \mathrm{c}^2\)
2 \(4 a^2 c^2\)
3 \(8 a^2 c^2\)
4 \(2 a^2 c^2\)
5 \(\mathrm{a}^2 \mathrm{c}^2\)
Sequence and Series

118493 If \(a_1, a_2, a_3, a_4\) are in \(A P\), then \(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\frac{1}{\sqrt{a_3}+\sqrt{a_4}}\) is equal to

1 \(\frac{\sqrt{a_4}-\sqrt{a_1}}{a_3-a_2}\)
2 \(\frac{\mathrm{a}_4-\mathrm{a}_1}{\mathrm{a}_3-\mathrm{a}_2}\)
3 \(\frac{\mathrm{a}_3-\mathrm{a}_2}{\sqrt{\mathrm{a}_4}-\sqrt{\mathrm{a}_1}}\)
4 \(\frac{\mathrm{a}_1-\mathrm{a}_4}{\mathrm{a}_3-\mathrm{a}_1}\)
5 \(\frac{a_5-a_0}{a_1-a_4}\)
Sequence and Series

118494 If the product of five consecutive terms of a GP is \(\frac{243}{32}\), then the middle term is

1 \(2 / 3\)
2 \(3 / 2\)
3 \(4 / 3\)
4 \(3 / 4\)
5 1
Sequence and Series

118495 If \(6^{\text {th }}\) term of a GP is 2 , then the product of first 11 terms of the GP is equal to

1 512
2 1024
3 2048
4 256
5 32
Sequence and Series

118496 If a,b,c are in \(A P\) and if their squares taken in the same order form a GP, then \((a+c)^4\) is equal to.

1 \(16 \mathrm{a}^2 \mathrm{c}^2\)
2 \(4 a^2 c^2\)
3 \(8 a^2 c^2\)
4 \(2 a^2 c^2\)
5 \(\mathrm{a}^2 \mathrm{c}^2\)
Sequence and Series

118493 If \(a_1, a_2, a_3, a_4\) are in \(A P\), then \(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\frac{1}{\sqrt{a_3}+\sqrt{a_4}}\) is equal to

1 \(\frac{\sqrt{a_4}-\sqrt{a_1}}{a_3-a_2}\)
2 \(\frac{\mathrm{a}_4-\mathrm{a}_1}{\mathrm{a}_3-\mathrm{a}_2}\)
3 \(\frac{\mathrm{a}_3-\mathrm{a}_2}{\sqrt{\mathrm{a}_4}-\sqrt{\mathrm{a}_1}}\)
4 \(\frac{\mathrm{a}_1-\mathrm{a}_4}{\mathrm{a}_3-\mathrm{a}_1}\)
5 \(\frac{a_5-a_0}{a_1-a_4}\)
Sequence and Series

118494 If the product of five consecutive terms of a GP is \(\frac{243}{32}\), then the middle term is

1 \(2 / 3\)
2 \(3 / 2\)
3 \(4 / 3\)
4 \(3 / 4\)
5 1
Sequence and Series

118495 If \(6^{\text {th }}\) term of a GP is 2 , then the product of first 11 terms of the GP is equal to

1 512
2 1024
3 2048
4 256
5 32
Sequence and Series

118496 If a,b,c are in \(A P\) and if their squares taken in the same order form a GP, then \((a+c)^4\) is equal to.

1 \(16 \mathrm{a}^2 \mathrm{c}^2\)
2 \(4 a^2 c^2\)
3 \(8 a^2 c^2\)
4 \(2 a^2 c^2\)
5 \(\mathrm{a}^2 \mathrm{c}^2\)
Sequence and Series

118493 If \(a_1, a_2, a_3, a_4\) are in \(A P\), then \(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\frac{1}{\sqrt{a_3}+\sqrt{a_4}}\) is equal to

1 \(\frac{\sqrt{a_4}-\sqrt{a_1}}{a_3-a_2}\)
2 \(\frac{\mathrm{a}_4-\mathrm{a}_1}{\mathrm{a}_3-\mathrm{a}_2}\)
3 \(\frac{\mathrm{a}_3-\mathrm{a}_2}{\sqrt{\mathrm{a}_4}-\sqrt{\mathrm{a}_1}}\)
4 \(\frac{\mathrm{a}_1-\mathrm{a}_4}{\mathrm{a}_3-\mathrm{a}_1}\)
5 \(\frac{a_5-a_0}{a_1-a_4}\)
Sequence and Series

118494 If the product of five consecutive terms of a GP is \(\frac{243}{32}\), then the middle term is

1 \(2 / 3\)
2 \(3 / 2\)
3 \(4 / 3\)
4 \(3 / 4\)
5 1
Sequence and Series

118495 If \(6^{\text {th }}\) term of a GP is 2 , then the product of first 11 terms of the GP is equal to

1 512
2 1024
3 2048
4 256
5 32
Sequence and Series

118496 If a,b,c are in \(A P\) and if their squares taken in the same order form a GP, then \((a+c)^4\) is equal to.

1 \(16 \mathrm{a}^2 \mathrm{c}^2\)
2 \(4 a^2 c^2\)
3 \(8 a^2 c^2\)
4 \(2 a^2 c^2\)
5 \(\mathrm{a}^2 \mathrm{c}^2\)