Question
How many terms are there in the A.P.?
$18, 15\frac{1}{2},13,\ ....,-47$
$18, 15\frac{1}{2},13,\ ....,-47$
✓
Answer
Given,
A.P. $18, 15\frac{1}{2},13,\ ....,-47$
Here,
First term $a = 18$
Difference $\text{d}=15\frac{1}{2}-18=\frac{-5}{2}$
Last $n^{th}$ term $a_n = -47$
We know, $n^{th}$ term of A.P.
$a_n = a + (n - 1)d$
$\Rightarrow\ -47=18+(\text{n}-1)\frac{-5}{2}$
$\Rightarrow\ -47=18\frac{-5\text{n}}{2}+\frac{5}{2}$
$\Rightarrow\ \frac{5\text{n}}{2}=18+47+\frac{5}{2}$
$\Rightarrow\ \frac{5\text{n}}{2}=\frac{36+94+5}{2}$
$\Rightarrow\ 5\text{n}=135$
$\Rightarrow\ \text{n}=\frac{135}{5}$
$\Rightarrow\ \text{n}=27$
Hence, Total $27$ terms in given A.P.
A.P. $18, 15\frac{1}{2},13,\ ....,-47$
Here,
First term $a = 18$
Difference $\text{d}=15\frac{1}{2}-18=\frac{-5}{2}$
Last $n^{th}$ term $a_n = -47$
We know, $n^{th}$ term of A.P.
$a_n = a + (n - 1)d$
$\Rightarrow\ -47=18+(\text{n}-1)\frac{-5}{2}$
$\Rightarrow\ -47=18\frac{-5\text{n}}{2}+\frac{5}{2}$
$\Rightarrow\ \frac{5\text{n}}{2}=18+47+\frac{5}{2}$
$\Rightarrow\ \frac{5\text{n}}{2}=\frac{36+94+5}{2}$
$\Rightarrow\ 5\text{n}=135$
$\Rightarrow\ \text{n}=\frac{135}{5}$
$\Rightarrow\ \text{n}=27$
Hence, Total $27$ terms in given A.P.
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