if the sum of first 31 terns of an AP is 500 find 5th term
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S31=31/2 {2a+(31-1)d}
sum of 31 term is 500 therefore
500=31/2 {2a+(31-1)d}
1000=31 {2a+30d}
1000=62 {a+15d}
500/31 =a+15d
500/31-15d=a........ (1)
for 5 th term
s5=5/2 {2a+(5-1d)}
S5=5/2 {2 (500/31-15d)+4d}
S5=5/2 {2 (500-465d)/31+4d }
S5=(2500-2325d +124d)/31
S5=(2500-2201)d
sum of 31 term is 500 therefore
500=31/2 {2a+(31-1)d}
1000=31 {2a+30d}
1000=62 {a+15d}
500/31 =a+15d
500/31-15d=a........ (1)
for 5 th term
s5=5/2 {2a+(5-1d)}
S5=5/2 {2 (500/31-15d)+4d}
S5=5/2 {2 (500-465d)/31+4d }
S5=(2500-2325d +124d)/31
S5=(2500-2201)d
Ruhi5884:
I didn't get the exact answer
solve thar
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??????????????????
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Thus, the sum of the first 31 terms of the A.P. is 1271/3
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