Question

The sum of first 9 terms and first 21 terms of an AP are 63 and 273 respectively then find the sum of first 50 terms​

Answer 1

Answer

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

for n= 9

$s_{9}$ =$\frac{9}{2}$ {2a + 8d} = 63.                   (1)

$s_{21}$ =$\frac{21}{2}$ {2a + 20d} = 273            (2)

eqn (1)

2a + 8d = 14                                (3)

eqn (2)

2a + 20d = 26                           (4)

eqn (4) - eq (3) we get

12 d =12

therefore d = 1

put d =1 in eqn (3)

2a + 8(1) =14

2a = 6

a = 3

for n=50

$s_{50}$ = $\frac{50}{2}${2a + (50-1)d}

$s_{50}$ =25 (2*3+ 49*1)

    =25*55

    = 1375

Step-by-step explanation: