if the 15th term of an ap is 3 more than twice its 7th term. if the 1pth term of the ap is 41. find the 20th term
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Answer:
a+14d=2(a+6d)+3
⇒a-2d+3=0 ...(i)
again
a+9d=41 (ii)
subtracting i from ii
11d=44
⇒d=4
hence from ii
a=5
hence nth term = 5+ (n-1) x4
=4n+1
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