Sum of first four terms of an A.P. is 56 . sum of last four terms is 112 . If its first term is 11 ,

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8. Sequences and Series

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The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$, the number of terms is

A

$10$

B

$11$

C

$12$

D

None of these

Solution

(b) $11 + (11 + d) + (11 + 2d) + (11 + 3d) = 56$

$ \Rightarrow $$d = 2$

and $\{ 11 + (n – 1)2\} + \{ 11 + (n – 2)2\} + \{ 11 + (n – 3)2\}$$ + \{ 11 + (n – 4)2\} = 112$

$ \Rightarrow $ $n = 11$.

Standard 11

Mathematics

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