Sum of first 20 terms of AP is 400 and sum of first 40 terms is 1600 find sum of first 10 terms
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Let the first term be a and common difference be d and sum of 20 terms be S20\xa0{tex}S _ { n } = \\frac { n } { 2 } ( 2 a + (n -1) d ){/tex}{tex}S _ { 20 } = \\frac { 20 } { 2 } ( 2 a + (20 -1) d ){/tex}{tex}S _ { 20 } = \\frac { 20 } { 2 } ( 2 a + 19 d ){/tex}or,\xa0{tex}400 = \\frac { 20 } { 2 } ( 2 a + 19 d ){/tex}or, 400 = 10 [2a + 19d]or 2a + 19d = 40 ………. (i)Also, {tex}S _ { 40 } = \\frac { 40 } { 2 } ( 2 a + (40 -1) d ){/tex}S40 = {tex}\\frac{40}{2}{/tex}(2a + 39d)or, 1600 = 20[2a + 39d]or 2a + 39d = 80 ……….. (ii)Subtract (i) and (ii), we get(2a + 39d) – (2a + 19d) = 80 – 402a + 39d – 2a – 19d = 4020d = 40d = 2Put d = 2 in (i)2a + 19d = 402a + 19(2) = 402a + 38 = 402a = 40 – 382a = 2a = 1a = 1 and d = 2{tex}S _ { 10 } = \\frac { 10 } { 2 } [ 2 \\times 1 + ( 10 – 1 ) ( 2 ) ]{/tex}= 5 [2 + 9 {tex}\\times{/tex}\xa02]= 5 [2 + 18]= 5 {tex}\\times{/tex}\xa020= 100