sum of n terms of 1st = (n/2)(10 + 36(n-1))
= (n/2)(36n - 26)
sum of 2n terms of 2nd = (2n/2)(72 + 5(2n-1))
= n(10n + 67)
(n/2)(36n - 26) = n(10n + 67)
divide by n, and multiply by 2
36n - 26 = 20n + 134
16n = 160
n = 10
check:
sum(10 for 1st) = 5(10 + 36(9)) = 1670
sum(20 for 2nd) = 10(72 + 19(5)) = 1670
all is good!
the sum of the n termsof an ap whose first term is 5 and common difference is 36 is equal to the sum of 2n terms of another AP whose first term is 36 ans common difference is 5. Find n?
9 answers
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