Q : 3   In an AP:

           (iv) given \small a_3=15, S_1_0=125, find \small d and \small S_1_0
.

Answers (1)

It is given that
\small a_3=15, S_1_0=125
a_{3}= a+2d = 15 \ \ \ \ \ \ \ \ -(i)
Now, we know that 
S_n = \frac{n}{2}\left \{ 2a+(n-1)d \right \}
\Rightarrow S_{10} = \frac{10}{2}\left \{ 2\times(a) +(10-1)d\right \}
\Rightarrow 125 = 5\left \{ 2a+9d\right \}
\Rightarrow 2a+9d = 25 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)
On solving equation (i) and (ii) we will get 
a= 17 \ and \ d = -1
Now,
a_{10} = a+ 9d = 17 + 9(-1)= 17-9 = 8
Therefore, the value of d and 10th terms is -1 and 8 respectively

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