Question
If $\frac{4}{5}$, $a, 2$ are three consecutive terms of an A.P.?
✓
Answer
Here, we are given three consecutive terms of an A.P.
First term $(a_1)$ = $\frac{4}{5}$
Second term $(a_2) = a$
Third term $(a_3) = 2$
We need to find the value of a. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
$d = a_2 - a_1$
$\text{d}=\text{a}-\frac{4}{5}\ ....\text{(i)}$
Also,
$d = a_3 - a_2$
$d = 2 - a$
Now, on equating (i) and (ii), we get,
$\text{a}-\frac{4}{5}=2-\text{a}$
$\text{a}+\text{a}=2+\frac{4}{5}$
$2\text{a}=\frac{10+4}{5}$
$\text{a}=\frac{14}{10}$
$\text{a}=\frac{7}{5}$
Therefore, $\text{a}=\frac{7}{5}$.
First term $(a_1)$ = $\frac{4}{5}$
Second term $(a_2) = a$
Third term $(a_3) = 2$
We need to find the value of a. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
$d = a_2 - a_1$
$\text{d}=\text{a}-\frac{4}{5}\ ....\text{(i)}$
Also,
$d = a_3 - a_2$
$d = 2 - a$
Now, on equating (i) and (ii), we get,
$\text{a}-\frac{4}{5}=2-\text{a}$
$\text{a}+\text{a}=2+\frac{4}{5}$
$2\text{a}=\frac{10+4}{5}$
$\text{a}=\frac{14}{10}$
$\text{a}=\frac{7}{5}$
Therefore, $\text{a}=\frac{7}{5}$.
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