Maths Q6 – JEE-MAIN 2026

Home Maths Sequences and SeriesQuestion #403

The sum of the first ten terms of an A.P. is

160

and the sum of the first two terms of a G.P. is

8

. If the first term of the A.P. is equal to the common ratio of the G.P. and the first term of the G.P. is equal to common difference of the A.P., then the sum of all possible values of the first term of the G.P. is:

Clue 1 of 3: Define variables and relationships

Clearly define the first term and common difference/ratio for both AP and GP, and then translate the given conditions into algebraic equations.

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Step 1: Set up equations based on given conditions✦ Active

Let the first term of the A.P. be $a$ and its common difference be $d$ . Let the first term of the G.P. be $A$ and its common ratio be $R$ . The given conditions are:

S_{10} = \frac{10}{2} (2 a + 9 d) = 160 ⟹ 5 (2 a + 9 d) = 160 ⟹ 2 a + 9 d = 32 (1)

S_{2}^{'} = A + A R = 8 ⟹ A (1 + R) = 8 (2)

Also, we are given the relationships: $a = R (3)$ and $A = d (4)$ .

Step 2: Substitute relationships to form a single quadratic equation○ Expand

Substitute equations (3) and (4) into equation (1):

2 R + 9 A = 32 (5)

From equation (2), express $R$ in terms of $A$ :

R = \frac{8}{A} - 1 = \frac{8 - A}{A}

(Note:

A \neq 0

because

A (1 + R) = 8

implies

A \neq 0

. Also, if

A = 8

, then

R = 0

, which leads to

2 (0) + 9 (8) = 32 ⟹ 72 = 32

, a contradiction. So

A \neq 8

and

R \neq 0

Substitute this expression for $R$ into equation (5):

2 (\frac{8 - A}{A}) + 9 A = 32

Multiply by $A$ to clear the denominator:

2 (8 - A) + 9 A^{2} = 32 A

Rearrange into a standard quadratic form:

16 - 2 A + 9 A^{2} = 32 A ⟹ 9 A^{2} - 34 A + 16 = 0

Step 3: Calculate the sum of possible values of the first term of the G.P.○ Expand

The equation $9 A^{2} - 34 A + 16 = 0$ is a quadratic equation for $A$ . The question asks for the sum of all possible values of the first term of the G.P., which are the roots of this quadratic equation.

For a quadratic equation $p x^{2} + q x + r = 0$ , the sum of the roots is given by $- \frac{q}{p}$ . In this case, $p = 9$ , $q = - 34$ , and $r = 16$ .

Sum of possible values of A = - \frac{(- 34)}{9} = \frac{34}{9}

💡 Teacher's Secret Hint

Remember Vieta's formulas for the sum and product of roots of a quadratic equation.

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Maths Question 6 – JEE-MAIN 2026

Clue 1 of 3: Define variables and relationships

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