JEE Main 2026 — Mathematics Algebra

Let the first term of the A.P. be $a$a and its common difference be $d$d.

Let the first term of the G.P. be $A$A and its common ratio be $R$R.

Given that the sum of the first ten terms of the A.P. is $160$160:

$$\dfrac{10}{2} [2a + (10 - 1)d] = 160$$210​[2a+(10−1)d]=160

$$5(2a + 9d) = 160$$5(2a+9d)=160

$$2a + 9d = 32$$2a+9d=32

Given that the sum of the first two terms of the G.P. is $8$8:

$$A + AR = 8$$A+AR=8

$$A(1 + R) = 8$$A(1+R)=8

We are also given that $a = R$a=R and $A = d$A=d. Substituting these into the first equation:

$$2R + 9A = 32$$2R+9A=32

$$R = \dfrac{32 - 9A}{2}$$R=232−9A​

Substitute $R$R into the second equation:

$$A \left(1 + \dfrac{32 - 9A}{2}\right) = 8$$A(1+232−9A​)=8

$$A \left(\dfrac{2 + 32 - 9A}{2}\right) = 8$$A(22+32−9A​)=8

$$A(34 - 9A) = 16$$A(34−9A)=16

$$34A - 9A^2 = 16$$34A−9A2=16

$$9A^2 - 34A + 16 = 0$$9A2−34A+16=0

This is a quadratic equation in $A$A. The discriminant is $\Delta = (-34)^2 - 4(9)(16) = 1156 - 576 = 580 > 0$Δ=(−34)2−4(9)(16)=1156−576=580>0, so there are two distinct real values for $A$A.

The sum of all possible values of the first term of the G.P. ($A$A) is the sum of the roots of this quadratic equation:

Sum of roots $= -\dfrac{\text{coefficient of } A}{\text{coefficient of } A^2} = \dfrac{34}{9}$=−coefficient of A2coefficient of A​=934​

Answer: $\dfrac{34}{9}$934​