If the sum of n terms of an A.P. is 3n^2 + 5n and its mth term is 164, find the value of m.
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ap find m
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3n2+5n=n/2(2a+(n-1)d)dtweah wrote:Choose Cmaihuna wrote:If the sum of n terms of an A.P. is 3n^2 + 5n and its mth term is 164, find the value of m.
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27
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6n2 +10n=(2a-d)n+n2d
Which means d=6
and 2a-d=10
a=8
8+(m-1)6=164
m=27
- Vemuri
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I am missing something & its getting late for me (gotto go to sleep). Just putting my thoughts.maihuna wrote:If the sum of n terms of an A.P. is 3n^2 + 5n and its mth term is 164, find the value of m.
13
23
27
33
39
Sum of n terms => n/2[2a+(n-1)d] = 3n^2+5n
nth term => a+(n-1)d
n - number of terms
a - first term
d - difference between the terms in the AP
mth term => a+(m-1)d = 164
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- sureshbala
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Givne Sn = 3n^2 + 5n.
S1 = 8 i.e. first term =8.
S2 = 3(4)+5(2) = 22.
So sum of the first 2 terms is 22.
We have first term = 8.
Hence second term = 14.
So common difference d = 14 - 8 = 6.
Now mth term = 8+(m-1)6 = 164
i.e. m-1 =26
i.e m =27
S1 = 8 i.e. first term =8.
S2 = 3(4)+5(2) = 22.
So sum of the first 2 terms is 22.
We have first term = 8.
Hence second term = 14.
So common difference d = 14 - 8 = 6.
Now mth term = 8+(m-1)6 = 164
i.e. m-1 =26
i.e m =27