Maths Q6 – JEE-MAIN 2025

Let

a_{n}

be the

n^{t h}

term of an A.P. If

S_{n} = a_{1} + a_{2} + a_{3} + . . . + a_{n} = 700

a_{6} = 7

and

S_{7} = 7

, then

a_{n}

is equal to :

P. properties. The problem involves the $n^{t h}$ term and the sum of $n$ terms of an Arithmetic Progression.

Step 1: Formulate equations for the first term (

a

) and common difference (

d

).✦ Active

Given $a_{6} = 7$ and $S_{7} = 7$ . Using the formula for the $k^{t h}$ term of an A.P., $a_{k} = a + (k - 1) d$ , we have:

a_{6} = a + 5 d = 7 (Equation 1)

Using the formula for the sum of the first $k$ terms of an A.P., $S_{k} = \frac{k}{2} (2 a + (k - 1) d)$ , we have:

S_{7} = \frac{7}{2} (2 a + 6 d) = 7 ⟹ 7 (a + 3 d) = 7 ⟹ a + 3 d = 1 (Equation 2)

Subtracting Equation 2 from Equation 1:

(a + 5 d) - (a + 3 d) = 7 - 1 ⟹ 2 d = 6 ⟹ d = 3

Substitute $d = 3$ into Equation 2:

a + 3 (3) = 1 ⟹ a + 9 = 1 ⟹ a = - 8

Step 2: Determine the number of terms (

n

).○ Expand

Given $S_{n} = 700$ . Using the sum formula with $a = - 8$ and $d = 3$ :

\frac{n}{2} (2 a + (n - 1) d) = 700

Substitute the values of $a$ and $d$ :

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

Simplify and solve the quadratic equation for $n$ :

n (- 16 + 3 n - 3) = 1400 ⟹ n (3 n - 19) = 1400 ⟹ 3 n^{2} - 19 n - 1400 = 0

Using the quadratic formula $n = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

n = \frac{19 \pm \sqrt{(- 19)^{2} - 4 (3) (- 1400)}}{2 (3)} = \frac{19 \pm \sqrt{361 + 16800}}{6} = \frac{19 \pm \sqrt{17161}}{6}

Since $\sqrt{17161} = 131$ and $n$ must be a positive integer:

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

Step 3: Calculate the

n^{t h}

term (

a_{n}

).○ Expand

We need to find $a_{n}$ , which is $a_{25}$ since $n = 25$ . Using the formula $a_{k} = a + (k - 1) d$ :

a_{25} = a + (25 - 1) d = a + 24 d

Substitute $a = - 8$ and $d = 3$ :

a_{25} = - 8 + 24 (3) = - 8 + 72 = 64

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Maths Question 6 – JEE-MAIN 2025