Q) The sum of first n terms of an A.P. is 2n ² + 13n. Find its n ^th term and hence 10 ^th term. 

(Q 29 B – 30/5/2 – CBSE 2026 Question Paper)

Ans:

Let’s consider the first term is a and Common difference is d

(i) n ^th term:

Method 1:

Step 1: n ^th term can be found out if we subtract sum of first (n-1) terms from sum of first n terms.

∴ T_n = S_n – S_n-1

Step 2: Now, sum of first n terms of an A.P. is 2n ² + 13n

∴ S_n = 2 n ² + 13 n = n (2 n + 13)

Step 3: sum of first (n-1) terms: S_n-1 = (n – 1)(2 (n – 1) + 13)

= (n – 1)(2n + 11) = 2 n ² +11 n – 2 n -11 = 2 n ² + 9 n -11

∴ T_n = (2 n ² + 13 n) – (2 n ² + 9 n – 11)

∴ T_n = 2 n ² + 13 n – 2 n ² – 9 n + 11

∴ T_n = 4 n + 11

Therefore, n ^th term of the AP is 4 n + 11.

Method 2:

Step 4: it is given that sum of first n terms of the AP is 2 n ² + 13 n.

∴ Sum of 1 term, S₁ = 2 (1) ² + 13 (1)

= 2 + 13 = 15

This becomes 1 ^st term of the AP, ∴ a = 15

Step 5: Similarly, sum of first 2 terms, S₂ = 2 (2) ² + 13 (2)

= 8 + 26 = 34

∵ 2 ^nd term, T₂ = S₂ – S₁ = 34 – 15 = 19

Now common difference of AP, d = T₂ – T₁

∴ d = 19 – 15 = 4

Now, we have first term of the AP as 15 and common differnece as 4.

Step 6: n ^th term of an AP is given by, T_n = a + (n – 1) d

= 15 + (n – 1) (4) = 15 + 4 n – 4

= 4 n + 11

Therefore, n ^th term of the AP is 4 n + 11.

(ii) 10 ^th term:

Step 4: Value of 10 ^th term, T₁₀ = 4 (10) + 11

∴ T₁₀ = 40 + 11 = 51

Therefore, value of 10 ^th term is 51.

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