# Question #ff2fc

Aug 31, 2016

$\text{ The Reqd. Sum} = \frac{580}{9} \approx 64.44$.

#### Explanation:

Let the A.P. be,

$a , a + d , a + 2 d , a + 3 d , \ldots , \text{where} , a , d , \in \mathbb{R} , \mathmr{and} , d \ne 0$.

Let ${t}_{n} , \mathmr{and} , {S}_{n}$ be the ${n}^{t h}$ term, &, the sum of first $n$

terms of A.P. resp.

We know that, ${t}_{n} = a + \left(n - 1\right) d , \mathmr{and} , {S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right]$.

${t}_{3} : {t}_{6} = 9 : 4 \ldots . \text{[Given]} \Rightarrow \frac{a + 2 d}{a + 5 d} = \frac{9}{4.} \ldots \ldots \ldots . \left(1\right)$

Further, $\text{Given } {S}_{5} = 60 \Rightarrow \frac{5}{2} \left(2 a + 4 d\right) = 60 , \mathmr{and} , \left(a + 2 d\right) = 12$.

By $\left(1\right) , \text{ then, } a + 5 d = \frac{4 \cdot 12}{9} = \frac{16}{3}$

Therefore, $\left(a + 2 d\right) - \left(a + 5 d\right) = 12 - \frac{16}{3} \Rightarrow - 3 d = \frac{20}{3} \Rightarrow d = - \frac{20}{9}$.

Since, $a + 2 d = 12$, we have,

$a = 12 - 2 \left(- \frac{20}{9}\right) = 12 + \frac{40}{9} = \frac{148}{9}$.

Finally, the Reqd. Sum $= {S}_{10} = \frac{10}{2} \left(2 a + 9 d\right)$

$= 5 \left[2 \cdot \frac{148}{9} + 9 \left(- \frac{20}{9}\right)\right]$

$= 5 \left[\frac{296}{9} - \frac{180}{9}\right]$

$= \frac{5 \cdot 116}{9}$

$= \frac{580}{9} \approx 64.44$.

Enjoy Maths.!