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Algebra Level 2

Given that the sum of the first n n terms of an arithmetic progression is given by 3 n 2 2 + 13 n 2 \dfrac{3n^2}2 + \dfrac{13n}2 , find the value of the 2 5 th 25^\text{th} term.


The answer is 80.

Anish Harsha by Anish Harsha , 20, India

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3 solutions

Using A.P.
⇒ n ( a + a n ) 2 = 3 n 2 + 13 n 2 \Rightarrow \dfrac{n(a+a_n)} {2}=\dfrac{3n^2+13n}{2}

25 ( 8 + a 25 ) = 25 ( 3 × 25 + 13 ) 25(8+a_{25})=25(3×25+13)

a 25 = 88 − 8 = 80 a_{25}=88-8=\boxed{80}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Hung Woei Neoh
May 30, 2016

T 25 = S 25 − S 24 = ( 3 ( 25 ) 2 2 + 13 ( 25 ) 2 ) − ( 3 ( 24 ) 2 2 + 13 ( 24 ) 2 ) = 1 2 ( 3 ( 25 ) 2 − 3 ( 24 ) 2 + 13 ( 25 ) − 13 ( 24 ) ) = 1 2 ( 3 ( 2 5 2 − 2 4 2 ) + 13 ( 25 − 24 ) ) = 1 2 ( 3 ( 25 − 24 ) ( 25 + 24 ) + 13 ) = 1 2 ( 3 ( 49 ) + 13 ) = 1 2 ( 147 + 13 ) = 160 2 = 80 T_{25} = S_{25} - S_{24}\\ =\left(\dfrac{3(25)^2}{2} + \dfrac{13(25)}{2}\right) - \left(\dfrac{3(24)^2}{2} + \dfrac{13(24)}{2}\right)\\ =\dfrac{1}{2}\left(3(25)^2 - 3(24)^2+13(25)-13(24)\right)\\ =\dfrac{1}{2}\left(3(25^2-24^2) + 13(25-24)\right)\\ =\dfrac{1}{2}\left(3(25-24)(25+24) + 13\right)\\ =\dfrac{1}{2}\left(3(49)+13\right)\\ =\dfrac{1}{2}\left(147+13\right)\\ =\dfrac{160}{2}\\ =\boxed{80}

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Denton Young
Aug 28, 2016

The first term is ( 3 ∗ 1 2 ) / 2 + ( 13 ∗ 1 ) / 2 (3 * 1^2)/2 + (13*1)/2 = 8

The sum of the first 2 terms is ( 3 ∗ 2 2 ) / 2 + ( 13 ∗ 2 ) / 2 (3 * 2^2)/2 + (13*2)/2 = 19

So the second term is 19 - 8 = 11

Difference is 11 - 8 = 3

25th term = 8 + (3 * (25-1)) = 80

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