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In an AP, given $a_{3} = 15, S_{10} = 125$ . Find d and $a_{10}$ .

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3, - 9

- 6, 10

5, 7

- 1, 8

answer is D.

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Detailed Solution

We know,

S_{n} = \frac{n}{2} [a + a_{n}]

And according to the question, we know the value of

S_{10}

,
So comparing, we get the value of n, i.e.,

n = 10

So, now putting the value of n in the above equation,

{\Rightarrow S}_{10} = \frac{10}{2} [a + a_{10}]

\Rightarrow 125 = \frac{10}{2} [a + a_{10}]

\Rightarrow 125 = 5 [a + a_{10}]

\Rightarrow \frac{125}{5} = a + a_{10}

\Rightarrow 25 = a + a_{10}

\Rightarrow a_{10} = 25 - a

[equation- I ]
Now,

a_{n} = a + (n - 1) d

\Rightarrow a_{10} = a + (10 - 1) d

\Rightarrow 25 - a = a + 9 d

[from equation- I )

\Rightarrow 25 = a + a + 9 d

\Rightarrow 25 = 2 a + 9 d

\Rightarrow 2 a + 9 d = 25

[equation -II ]
Now,
Given:

a_{3} = 15

\Rightarrow a + (3 - 1) d = 15

\Rightarrow a + 2 d = 15

[equation- III ] Multiplying equation- III with 2 and equation- II with 1
After multiplying, the equations are:

\Rightarrow 2 a + 4 d = 30

\Rightarrow 2 a + 9 d = 25

Subtracting both the above equations, we get,

\Rightarrow - 5 d = 5

\Rightarrow d = \frac{- 5}{5}

\Rightarrow d = - 1

Now, put the value of d in equation III,

a + 2 d = 15

\Rightarrow a + 2 \times (- 1) = 15

\Rightarrow a - 2 = 15

\Rightarrow a = 15 + 2

\Rightarrow a = 17

Now, to find the value of

a_{10}

, we will put the value of a and d found above, in the equation,

a_{n} = a + (n - 1) d

\Rightarrow a_{10} = 17 + (10 - 1) \times (- 1)

\Rightarrow a_{10} = 17 - 9

\Rightarrow a_{10} = 8

Therefore, the value of

d = - 1

and

a_{10} = 8

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