Question

if the sum of first 9 term of an ap is equal to the sum of first 11 terms then what is the sum of its first 20 terms

Answer 1

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Answer 2

Sum of first 20 terms = 0

Step-by-step explanation:

Given,

Sum of 9 first terms = Sum of first 11 terms

=> We know that sum of n terms of an AP can be calculated as :

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

- According to the given condition in the question ,

$S_{9} =S_{11}$

=> Calculating the values of  $S_{9} \ and \ S_{11}$

$S_{9} = \frac{9}{2}(2a+(9-1)d)\\\\ S_{9} = \frac{9}{2} (2a + 8d)----(i)$

$S_{11} = \frac{11}{2}(2a+(11-1)d)\\\\ S_{11} = \frac{11}{2} (2a + 10d)----(ii)$

=> Comparing equations (i) and (ii),

$\frac{9}{2}(2a+8d) = \frac{11}{2}(2a+10d)\\\\ 9(2a+8d)=11(2a+10d)\\18a + 72d = 22a + 110d\\22a-18a+110d-72d = 0\\4a + 38d = 0\\2a + 19d =0 ---(iii)$

=> Sum of first 20 terms will be ,

$S_{20} = \frac{20}{2}(2a+(20-1)d)\\\\ S_{20} = 10 (2a + 19d)$

(We know that 2a+19d =0 {from equation (iii)})

On substituing the value of 2a + 19d we will get $S_{20} = 0$.

=> Hence sum of first 20 terms = 0 .