Given $a_{3} = 15, S_{10} = 125$ , Find d and $a_{10}$

18

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4

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8

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6

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Solution

The correct option is D $8$
As we know that the nth term, $a_{n} = a + (n - 1) d$
& Sum of first $n$ terms, $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$ , where $a$ & $d$ are the first term & common difference of an AP.
$S_{10} = 125 = \frac{10}{2} (2 a + 9 d)$
$\Rightarrow 2 a + 9 d = 25$ ............... (i)
$a_{3} = a + 2 d = 15$ ............. (ii)
Subtracting equation (ii) from (i) we get
$a + 7 d = 10$ (the $8^{t h}$ term)
Subtracting $3^{r d}$ term from $8^{t h}$ term we get
$(a + 7 d) - (a + 2 d) = 10 - 15 = - 5$
$\Rightarrow 5 d = - 5$
$\Rightarrow d = - 1$
So, $a = 17$
$a_{10} = 17 - 9 = 8$

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Similar questions

Given , $a_{3} = 15, s_{10} = 125$ find $d$ and $a_{10}$ .

If $a_{3} = 15, S_{10} = 125$ , find $d$ and $a_{10}$

a_{3} = 15, S_{10} = 125

d

a_{10}

a_{3} = 15, S_{10} = 125

a_{10}

Question 3 (iv)
In an AP:
(iv) Given $a_{3} = 15, S_{10} = 125$ , find d and $a_{10}$ .

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