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8. Sequences and Series
easy
The solution of the equation $(x + 1) + (x + 4) + (x + 7) + ......... + (x + 28) = 155$ is
A
$1$
B
$2$
C
$3$
D
$4$
Solution
(a) We have $(x + 1) + (x + 4) + ……… + (x + 28) = 155$
Let $n$ be the number of terms in the $A.P.$ on $L.H.S.$
Then $x + 28 = (x + 1) + (n – 1)3$
$ \Rightarrow $ $n = 10$
$\therefore $ $(x + 1) + (x + 4) + …… + (x + 28) = 155$
$ \Rightarrow $ $\frac{{10}}{2}[(x + 1) + (x + 28)] = 155$
$ \Rightarrow $ $x = 1$.
Standard 11
Mathematics