JEE Main 2025 — Sequences And Series Question with Solution

Sum of n terms of an AP ($S_n$): $$S_n=\frac{n}{2}[2a+(n-1)d]$$ where 'a' is the first term and 'd' is the common difference.

nth term of an AP ($a_n$): $$a_n=a+(n-1)d$$

Given Information:

$S_n=700$ (The sum of the first n terms is 700)

$a_6=7$ (The 6th term is 7)

$S_7=7$ (The sum of the first 7 terms is 7)

Steps to Solve:

Use $S_7$ to find a relationship between 'a' and 'd':

$$S_7=\frac{7}{2}[2a+(7-1)d]=7$$

$$\frac{7}{2}[2a+6d]=7$$

$$2a+6d=2$$

$$a+3d=1$$ (Equation 1)

Use $a_6$ to find another relationship between 'a' and 'd':

$$a_6=a+(6-1)d=7$$

$$a+5d=7$$ (Equation 2)

Solve the system of equations (Equation 1 and Equation 2) to find 'a' and 'd':

Subtract Equation 1 from Equation 2:

$$(a+5d)-(a+3d)=7-1$$

$$2d=6$$

$$d=3$$

Substitute d = 3 into Equation 1:

$$a+3(3)=1$$

$$a+9=1$$

$$a=-8$$

Find 'n' using $S_n=700$:

$$S_n=\frac{n}{2}[2a+(n-1)d]=700$$

$$\frac{n}{2}[2(-8)+(n-1)(3)]=700$$

$$n[-16+3n-3]=1400$$

$$n[3n-19]=1400$$

$$3n^2-19n-1400=0$$

Solve the quadratic equation for 'n':

We can use the quadratic formula: $$n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$n=\frac{19\pm \sqrt{(-19{)}^2-4(3)(-1400)}}{2(3)}$$

$$n=\frac{19\pm \sqrt{361+16800}}{6}$$

$$n=\frac{19\pm \sqrt{17161}}{6}$$

$$n=\frac{19\pm 131}{6}$$

We have two possible values for n:

$$n=\frac{19+131}{6}=\frac{150}{6}=25$$

$$n=\frac{19-131}{6}=\frac{-112}{6}=-\frac{56}{3}$$ (Since 'n' must be a positive integer, we discard this solution)

So, $$n=25$$

Find $a_n$ (which is $a_{25}$):

$$a_{25}=a+(25-1)d$$

$$a_{25}=-8+(24)(3)$$

$$a_{25}=-8+72$$

$$a_{25}=64$$

Therefore, $a_n=64$. The correct answer is Option D.