Let be an A.P. and be an increasing G.P. If and , then is equal to: [2026]

Question

Let $a_{1}, a_{2}, a_{3}, . . .$ be an A.P. and $g_{1} = a_{1}, g_{2}, g_{3}, . . .$ be an increasing G.P. If $a_{1} = a_{2} + g_{2} = 1$ and $a_{3} + g_{3} = 4$ , then $a_{10} + g_{5}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(4)

$a_{1}, a_{2}, a_{3}, . . .$ be an A.P.and $g_{1}, g_{2}, . . .$ be a G.P.

$g_{2} = g_{1} r = a_{1} r$ ( $∵ g_{1} = a_{1}$ )

$\Rightarrow g_{2} = r$ ( $∵ a_{1} = 1$ )

$g_{3} = a_{1} r^{2} = r^{2}$

Now, $a_{1} = a_{2} + g_{2} = 1$

$\Rightarrow (1 + d) + r = 1$ [ $∵ a_{2} = a_{1} + d = 1 + d and g_{2} = r$ ]

$\Rightarrow d + r = 0 \Rightarrow d = - r$

Also, $a_{3} + g_{3} = 4$

$(1 + 2 d) + (r^{2}) = 4$

$\Rightarrow 1 - 2 r + r^{2} = 4 \Rightarrow r^{2} - 2 r - 3 = 0$

$\Rightarrow (r - 3) (r + 1) = 0$

$\Rightarrow r = 3$ (Neglecting r = –1 as it is an increasing G.P.)

$\Rightarrow d = - 3$

Now, $a_{10} + g_{5} = (1 + 9 d) + (r^{4})$

$= 1 + 9 (- 3) + (3^{4}) = 8 - 27 = 55$

Solution

Q.
Let $a_{1}, a_{2}, a_{3}, . . .$ be an A.P. and $g_{1} = a_{1}, g_{2}, g_{3}, . . .$ be an increasing G.P. If $a_{1} = a_{2} + g_{2} = 1$ and $a_{3} + g_{3} = 4$ , then $a_{10} + g_{5}$ is equal to: [2026]

1 81

2 76

3 62

4 55

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