The Sum of the first term of an Ap is given by sn=2nsquare+8n find the 16th term of the AP
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given
sn = 2n^2+8n
taking n=1
s1= 2×(1)^2+8×1
= 2×1+8
= 10
so sum of the first term of AP is 10.
bt sum of the first term will be first term.
so first term = 10
taking n=2.
2n^2+8n
= 2×(2)^2+8×2
= 2×4+16
= 24.
sum of the second term is 24
first term+ second term =24
10+ a2 = 24
a2= 24-10
a2= 14...
now common difference d= 2nd term - 1st term
= 14 -10
= 4
we need to find 16th term
s16 = a+ (16-1)d
= 10 + 15× 4
= 10+60
= 70 ...
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sn = 2n^2+8n
taking n=1
s1= 2×(1)^2+8×1
= 2×1+8
= 10
so sum of the first term of AP is 10.
bt sum of the first term will be first term.
so first term = 10
taking n=2.
2n^2+8n
= 2×(2)^2+8×2
= 2×4+16
= 24.
sum of the second term is 24
first term+ second term =24
10+ a2 = 24
a2= 24-10
a2= 14...
now common difference d= 2nd term - 1st term
= 14 -10
= 4
we need to find 16th term
s16 = a+ (16-1)d
= 10 + 15× 4
= 10+60
= 70 ...
hope this will help you dear ...
plz mark me as brainliest and follow me..
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