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Find the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77

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#### Solution

Sum of the first n terms of an Arithmetic progression is S_{n} = [2a + (n – 1)d]

Given S_{10} = 52

52 = `10/2`[ 2a + (10 – 1)d]

52 = 5[2a + 9d]

52 = 10a + 45d ......(1)

Also given S_{15} = 77

77 = 152[2a + (15 – 1)d]

77 = 152[2a + 14d]

77 = 15[a + 7d]

77 = 15a + 105d .......(2)

(1) × 15 ⇒ 780 = 150a + 675d

(2) × 10 ⇒ 770 = 150a + 1050d

10 = 0 – 375d

d = `10/(- 375)`

= `- 2/75`

Substituting the value of d in (1)

52 = `10"a" + 45(- 2/75)`

52 = `10"a" - 6/5`

10a = `52 + 6/5`

= `(260 + 6)/5`

= `266/5`

a = `266/(5 xx 10)`

= `133/25`

S^{20} = `20/2[2 xx 133/25 + 19 xx (-2)/75]`

= `10[266/25 - 18/75]`

= `10[(798- 38)/75]`

= `10[760/75]`

S^{20} = `2 xx 760/15`

= `2 xx 152/3`

= `304/3`

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