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Arithmetic Progressions — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsArithmetic Progressions5 Marks
Question
Let there be an A.P. with first term ' $a$ ', common difference ' $d$ '. If $a_n$ denotes in $n^{\text {th }}$ term and $S_n$ the sum of first $n$ terms, find. n and d , if $a =8, a _{ n }=62$ and $S _{ n }=210$.
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Answer

Here, we have an A.P. Whose $n^{\text {th }}$ term $\left(a_n\right)$, Sum of first $n$ terms $\left(S_n\right)$ and first term $(a)$ are given. We need to find the number of terms ( n ) and the common difference (d).
Here,
First term $(a) = 8$
Last term $(a_n) = 62$
Sum of n terms $(S_n) = 210$
Now, here the sum of the n terms is given by the formula,
$\text{S}_\text{n}=\Big(\frac{\text{n}}{2}\Big)(\text{a}+\text{l})$
Where, a = the first term
l = the last term
So, for the given A.P. on substituting the values in the formula for the sum of n terms of an A.P., we get,
$210=\Big(\frac{\text{n}}{2}\Big)[8+62]$
$210(2)=\text{n}(70)$
$\text{n}=\frac{420}{70}$
$\text{n}=6$
Also, here we will find the value of d using the formula.
$a_n = a + (n - 1)d$
So, Substituting the values in the above mentioned formula
$62 = 8 + (6 - 1)d$
$62 - 8 = (5)d$
$\frac{54}{5}=\text{d}$
$\text{d}=\frac{54}{5}$
Therefore, for the given A.P. n = 6 and $\text{d}=\frac{54}{5}$.

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