Question

The sum of first five terms of an ap are 35 and the sum of next three terms is 57 . find the ap and sum of first 35 numbers

Answer 1

Answer:

given S5 = 35 and S8 = 92

to find AP and S35

Sn = n/2[2a + (n-1)d]

now, 35 = 5/2(2a + 4d)

70 = 10a + 20 d

or a + 2d = 7 [ divide through by 10] .......1

again, 92 = 8/2[2a + 7d]

184 = 16a + 56d [ divide through by 8]

23 = 2a + 7d....................................................2

a + 2d = 7

2a + 7d = 23

on solving

d = 3 and a = 1

AP will be 1, 4, 7, 10.........Answer

now S35 = 35/2 [ 2*1 + 34*3]

= 35/2 [104]

= 1820 ............. Answer

now

Answer 2

$\star$  Question : The sum of first five terms of an ap are 35 and the sum of next three terms is 57 . Find the ap and sum of first 35 numbers

$\star$ Answer:

$\star$ 1 ) Series is 1 , 4 , 7 , 10 , 13 , 16 , 19 , 22 ,...

$\star$ 2 )  $S_{35}=1820$

$\star$ Step-by-step explanation:

$\star$ The sum of first five terms of an AP = $S_{5}=\frac{5}{2}*[2a+(n-1)d]=35$

$=>2a+4d=14\\\\=>a+2d=7...(1)$

$\star$ and The sum of next three terms is 57.

$=>S_8-S_5=57\\\\=>\frac{8}{2}*[2a+(8-1)d] -(\frac{5}{2}*[ 2a+(5-1)d])=57\\\\=>8a+28d-(5a+10d)=57\\\\=>8a+28d-5a-10d=57\\\\=>3a+18d=57\\\\=>a+6d=19...(2)$

$\star$ Now , Solve  (2) - (1) , we get ,

$(a+6d)-(a+2d)=19-7\\\\=>4d=12\\\\=>d=3$

$\star$ Now , sub. d=3 in (1) , we get ,

$=>a+2(3)=7\\\\=>a=7-6\\\\=>a=1$

$\star$ So, the AP is a , a+d , a+2d , a+3d , a+4d , a+5d , ......

$=>1,1+3,1+6,1+9,1+12,...\\\\=>1,4,7,10,13,16,19,...$

is required AP.

$Now,S_{35}=\frac{35}{2}*[2(1)+(35-1)3]\\\\ =>S_{35}=\frac{35}{2}*[2+102]\\\\=>S_{35}=\frac{35}{2}*[104]\\\\=>S_{35}=1820$

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