An important question from ML Aggarwal(avichal Publication). We have been asked to find the value of X.
Solve the equation : 1+4+7+10+…+x=287
Arithmetic Progression, Chapter 9, Question no 6
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Here,givena=1
d=4−1=3
and,sn=287
Now,
sn=2n(2a+(n−1)d)
⇒287=2n(2×1+(n−1)3)
⇒287=2n(2+3n−3)
⇒574=n(3n−1)
⇒574=3n2−n⇒3n2−n−574=0
on solving the quadratic equaton using formula n=2a−b±b2−4ac
We get n=14 & 3−41[does not exist] so, n=14
Now,
sn=2n(a+1)
⇒287=214(1+x)
⇒574=14(1+x)
⇒(1+x)=14574⇒1+x=41
⇒x=41−1
∴x=40
x=40 is the solution.