Question

If the sum of first \(n\) terms of an A.P. is \( \frac{3n^2}{2} + \frac{5n}{2} \) then its 25th term is :

• A. 70
• B. \(-n\)
• C. 76
• D. none of these

Hint:

Method 1: \(a_n = S_n - S_{n-1}\). \(S_n = \frac{3n^2+5n}{2}\). \(a_{25} = S_{25} - S_{24}\). \(S_{25} = (3 \cdot 25^2 + 5 \cdot 25)/2 = (1875 + 125)/2 = 2000/2 = 1000\). \(S_{24} = (3 \cdot 24^2 + 5 \cdot 24)/2 = (1728 + 120)/2 = 1848/2 = 924\). \(a_{25} = 1000 - 924 = 76\). Method 2: If \(S_n = An^2+Bn\), then \(a_1=A+B\) and common difference \(d=2A\). Here \(A=3/2, B=5/2\). \(a_1 = 3/2 + 5/2 = 8/2 = 4\). \(d = 2(3/2) = 3\). \(a_{25} = a_1 + (25-1)d = 4 + 24(3) = 4 + 72 = 76\).

Answer 1

The Correct Option is C

Concept: Let \(S_n\) be the sum of the first \(n\) terms of an A.P. Let \(a_n\) be the \(n^{th}\) term of the A.P. The \(n^{th}\) term can be found using the relation: \(a_n = S_n - S_{n-1}\) (for \(n>1\)). The first term \(a_1 = S_1\). Alternatively, if \(S_n\) is a quadratic in \(n\) of the form \(An^2+Bn\), then it represents the sum of an AP. The common difference \(d = 2A\) and the first term \(a_1 = A+B\). Then \(a_k = a_1 + (k-1)d\). Method 1: Using \(a_n = S_n - S_{n-1}\) Given \(S_n = \frac{3n^2}{2} + \frac{5n}{2}\). We need the 25th term, \(a_{25}\). So, \(a_{25} = S_{25} - S_{24}\). Calculate \(S_{25}\): \(S_{25} = \frac{3(25)^2}{2} + \frac{5(25)}{2} = \frac{3(625)}{2} + \frac{125}{2} = \frac{1875 + 125}{2} = \frac{2000}{2} = 1000\). Calculate \(S_{24}\): \(S_{24} = \frac{3(24)^2}{2} + \frac{5(24)}{2} = \frac{3(576)}{2} + \frac{120}{2} = \frac{1728 + 120}{2} = \frac{1848}{2} = 924\). Now, calculate \(a_{25}\): \(a_{25} = S_{25} - S_{24} = 1000 - 924 = 76\). Method 2: Finding \(a_1\) and \(d\) Given \(S_n = \frac{3}{2}n^2 + \frac{5}{2}n\). This is in the form \(An^2+Bn\) where \(A = \frac{3}{2}\) and \(B = \frac{5}{2}\). First term, \(a_1 = S_1 = \frac{3(1)^2}{2} + \frac{5(1)}{2} = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4\). (Using shortcut: \(a_1 = A+B = \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4\)). Common difference, \(d = 2A = 2 \times \frac{3}{2} = 3\). The \(k^{th}\) term of an A.P. is \(a_k = a_1 + (k-1)d\). We need the 25th term (\(k=25\)): \(a_{25} = a_1 + (25-1)d\) \(a_{25} = 4 + (24)(3)\) \(a_{25} = 4 + 72\) \(a_{25} = 76\). Both methods yield 76.