JEE Mathematics 2025 Question with Solution

Given: A G.P. $a_1, a_2, a_3, \ldots$ of increasing positive terms with

and

Find: The value of $a_6$.

Let the first term be $a$ and common ratio be $r$. Then

Using $a_1 a_5 = 28$,

so

Using $a_2 + a_4 = 29$,

so

Squaring this relation,

Divide equation $\text{(2)}$ by equation $\text{(1)}$:

which gives

Hence,

Since the terms are increasing positive terms, $r>1$, so we take the positive value and obtain

Now use equation $\text{(1)}$:

Substituting $r^2=28$,

so

and therefore

Now calculate $a_6$:

Substituting $a=\frac{1}{\sqrt{28}}$ and $r=\sqrt{28}$,

Therefore, the value of $a_6$ is $784$. The correct option is C.

Given: A G.P. $a_1, a_2, a_3, \ldots$ of increasing positive terms with

Find: The value of $a_6$.

Write the terms as

From $a_1 a_5=28$,

From $a_2+a_4=29$,

From equation $\text{(1)}$,

Substitute into equation $\text{(2)}$:

so

Hence,

which gives

Using the quadratic formula,

Now,

and

Therefore,

Since the terms are increasing, this is the valid root.

Now from equation $\text{(1)}$,

so

Finally,

Therefore, the value of $a_6$ is $784$. The correct option is C.