Question

The sum of first 5 terms of an ap is 55 and sum of ist 10 terms of the ap is 235 , find the sum of ist 20 terms

Answer 1

the sum of first 5 terms of an ap is 55

first five terms of AP are a, (a+d), (a+2d), (a+3d), (a+4d)

so, a + (a+d) + (a+2d) + (a+3d) + (a+4d) = 55
5a + 10d = 55
5(a + 2d) = 55
a + 2d = 55/5
a + 2d = 11................(1)

where a = 11 - 2d

sum of it's first 10 terms of AP is 235

first 10 terms of AP are a, (a+d), (a+2d), (a+3d), (a+4d), (a+5d), (a+6d), (a+7d), (a+8d), (a+9d)

so, a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d)+(a+6d)+(a+7d)+(a+8d)+(a+9d) = 235

10a + 45d = 235
5(2a + 9d) = 235
2a + 9d = 235/5
2a + 9d = 47
2(11 - 2d) + 9d = 47
22 - 4d + 9d = 47
5d = 47 - 22
5d = 25
d = 25/5
d = 5

a = 11 - 2d
a = 11 - 2(5)
a = 11 - 10
a = 1

sum of first 20 terms,
$S(n) = \frac{n}{2} (2a + (n - 1)d) \\ \\ where \: n = 20 \\a = 1 \\ d = 5 \\ S(20) = \frac{20}{2} (2 \times 1 + (20 - 1)5) \\ S(20) = 10(2 + 19 \times 5) \\ S(20) = 10(2 + 95) \\ S(20) = 10(97) \\ S(20) = 970$

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Answer 2

See the attachment

Solved using Sum of n terms formula

Sn = (n/2)×[2a + (n-1)d]

Hope this Helps