if the first term of an AP is 5 and the sum of first five terms is equal to one fourth of the sum of the next five terms‚ find the sum of first 40 terms
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sn= n/2 (2a+ (n-1)d)
a=5 n=5
s5= 5/2 (2(5)+(5-1)d)
s5= 5/2 ( 10+4d)=1/4[10/2 (10+9d)-5/2 (10+4d)]
25+10d= 50+45d-25+10d
15d= 75
d= 5
s40= 40/2 [2(5)+(40-1)5]
= 20[10+ 39(5)]
=205(20)
= 4100
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