Question

if (x+1) + (x+4) + (x+7) +........+ (x+28)=155, find x

Answer 1

x = 1........... . ......

Answer 2

Hey There!!

Here, we have:

(x+1) + (x+4) + (x+7) + ... + (x+28) = 155

We can see that it is an AP, with

First Term = a = x+1

Common Difference = d = 3

Let there be n terms.

$T_n=a+(n-1)d \\ \\ \implies x+28=(x+1)+3n-3 \\ \\ \implies 28=3n-2 \\ \\ \implies 3n=30 \\ \\ \implies \boxed{n=10}$


Now, Sum of n terms is given as:

$S_n=\frac{n}{2}(\text{First Term + Last Term}) \\ \\ \\ \implies S = \frac{10}{2} (x+1+x+28) \\ \\ \\ \implies 155 = 5(2x+29) \\ \\ \\ \implies 2x+29=\frac{155}{5} \\ \\ \\ \implies 2x+29=31 \\ \\ \implies 2x=2 \\ \\ \\ \implies \boxed{x=1}$


Hope it helps
Purva
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