Sequences and Series jee mains questions | Sequences and Series Previous Year Question

Q 31 :

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

760
755
750
757

1

2

3

4

(4)

We have, $a r + a r^{3} + a r^{5} = 21, a r^{7} + a r^{9} + a r^{11} = 15309$

$\Rightarrow a r (1 + r^{2} + r^{4}) = 21$ ... (i)

and $a r^{7} (1 + r^{2} + r^{4}) = 15309$ ... (ii)

Divide equation (ii) by (i), we get

$\Rightarrow \frac{a . r^{7}}{a r} = \frac{15309}{21} \Rightarrow r^{6} = 729$

$\Rightarrow r = 3 and a = \frac{7}{91}$

Now, $S_{9} = \frac{a \cdot (r^{9} - 1)}{r - 1}$

$= \frac{\frac{7}{91} (19683 - 1)}{2}$

$= \frac{7 \times 19682}{91 \times 2}$

$= \frac{9841}{13} = 757$ .

Q 32 :

Let $a_{1}, a_{2}, a_{3}, . . . . .$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ , then $a_{6}$ is equal to : [2025]

812
526
784
628

1

2

3

4

(3)

Here, r > 0 and $a_{1}, a_{2}, a_{3}, . . . . . > 0$ as G.P. has increasing positive terms.

$a_{1} a_{5} = 28$ (Given)

$\Rightarrow a (a r^{4}) = 28 = a^{2} r^{4} = 28$ ... (i)

Also, $a_{2} + a_{4} = 29$

$\Rightarrow a r + a r^{3} = 29$

$\Rightarrow a r (1 + r^{2}) = 29$ ... (ii)

From (i) and (ii), we get

$\frac{r^{2}}{{(1 + r^{2})}^{2}} = \frac{28}{{(29)}^{2}}$

$\Rightarrow 841 r^{2} = 28 r^{4} + 56 r^{2} + 28$

$\Rightarrow 28 r^{4} - 785 r^{2} + 28 = 0$

$\Rightarrow (r^{2} - 28) (28 r^{2} - 1) = 0$

$\Rightarrow r^{2} = 28 or \frac{1}{28}$

$\Rightarrow r = \sqrt{28}$ $[∵ r \neq \frac{1}{28} as G . P . is increasing]$

$\Rightarrow a = \frac{1}{\sqrt{28}} and a_{6} = a r^{5} = \frac{1}{\sqrt{28}} \times {(\sqrt{28})}^{5} = 784$ .

Q 33 :

Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]

$3 a_{99} - 100$
$3 a_{100} - 100$
$3 a_{100} + 100$
$3 a_{99} + 100$

1

2

3

4

(2)

We have, $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$

Let $2 x^{2} = 5 x - 3$

$\Rightarrow 2 x^{2} - 5 x + 3 = 0 \Rightarrow (2 x - 3) (x - 1) = 0$

$\Rightarrow x = 1, \frac{3}{2} ∴ a_{n} = A {(1)}^{n} + B {(\frac{3}{2})}^{n}$

Put n = 0; 0 = A + B

$n = 1; \frac{1}{2} = A + \frac{3}{2} B \Rightarrow B = 1, A = - 1$

$\Rightarrow a_{n} = (- 1) + {(\frac{3}{2})}^{n}$

$\Rightarrow \sum_{k = 1}^{100} a_{k} = \sum_{k = 1}^{100} (- 1) + {(\frac{3}{2})}^{k}$

$= - 100 + \frac{(\frac{3}{2}) ({(\frac{3}{2})}^{100} - 1)}{(\frac{3}{2} - 1)}$

$\Rightarrow - 100 + [3 ({(\frac{3}{2})}^{100} - 1)]$

$= 3 a_{100} - 100$ .

Q 34 :

Let the coefficients of three consecutive terms, $T_{r}, T_{r + 1}$ and $T_{r + 2}$ in the binomial expansion of ${(a + b)}^{12}$ be in a G.P. and let p be the number of all possible values or r. Let q be the sum of all rational terms in the binomial expansion of ${(\sqrt[4]{3} + \sqrt[3]{4})}^{12}$ . Then p + q is equal to : [2025]

299
287
295
283

1

2

3

4

(4)

Since, $T_{r}, T_{r + 1}$ and $T_{r + 2}$ are in G.P.

So, $\frac{T_{r + 1}}{T_{r}} = \frac{T_{r + 2}}{T_{r + 1}}$

$\Rightarrow \frac{{}^{12}C_{r}}{{}^{12}C_{r - 1}} = \frac{{}^{12}C_{r + 1}}{{}^{12}C_{r}} \Rightarrow \frac{13 - r}{r} = \frac{12 - r}{r + 1}$

$\Rightarrow 12 r - r^{2} = 13 r + 13 - r^{2} - r \Rightarrow 13 = 0$ (not possible)

So, p = 0

Now, ${(3^{1 / 4} + 4^{1 / 3})}^{12} = {}^{12}C_{0} {(3^{1 / 4})}^{12} {(4^{1 / 3})}^{0} + {}^{12}C_{12} {(3^{1 / 4})}^{0} {(4^{1 / 3})}^{12}$

$3^{3} + 4^{4} = 27 + 256 = 283 ∴ q = 283$

Thus, p + q = 0 + 283 = 283.

Topic Question Set

Q 31 :

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

Q 32 :

Let $a_{1}, a_{2}, a_{3}, . . . . .$ be a G.P. of increasing positive terms. If $a_{1} a_{5} = 28$ and $a_{2} + a_{4} = 29$ , then $a_{6}$ is equal to : [2025]

Q 33 :

Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]

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