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The sum of the first $9$ terms of an $AP$ is $81$ and the sum of its first $20$ terms is $400 .$ Find the first term, the common difference and the sum up to $15 th$ term.

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a = 1, d = 2,

S_{15} = 235

a = 3, d = 4, S_{15} = 215

a = 5, d = 3, S_{15} = 205

None of these

answer is D.

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Detailed Solution

Given, the sum of the first

9

terms of an

AP

81

and the sum of its first

20

terms is

400 .

Sum of n terms is given by,

S_{n} = \frac{n}{2} [2 \times a + (n - 1) d]

The given sum of

9

terms

= 81

\Rightarrow 81 = \frac{9}{2} [2 a + (9 - 1) d]

\Rightarrow 81 = \frac{9}{2} [2 a + 8 d]

\Rightarrow 162 = 18 a + 72 d . . . (1)

Sum of

20

terms

= 400

.
⇒

400 = \frac{20}{2} [2 a + (20 - 1) d]

\Rightarrow 400 = \frac{9}{2} [2 a + 19 d]

\Rightarrow 400 = 20 a + 190 d . . . . (2)

Multiplying

(1)

20

and

(2)

18

and then subtracting both the equation as below,

\Rightarrow 360 a + 3420 d = 7200

\Rightarrow 360 a + 1440 d = 3240

⇒

1980 d = 3690

\Rightarrow d = 2

Substituting the value of

d = 2

(2)

, we get,

\Rightarrow 400 = 20 a + 190 \times 2

\Rightarrow 400 = 20 a + 380

\Rightarrow 20 a = 20

\Rightarrow a = 1

Now the sum up to 15th term can be given as follow,

S_{15} = \frac{15}{2} [2 \times 1 + (15 - 1) 2]

⇒

S_{15} = \frac{15}{2} [2 + (14) 2]

⇒

S_{15} = \frac{15}{2} [2 + 28]

⇒

S_{15} = \frac{15}{2} [30]

⇒

S_{15} = 15 \times 15 = 225

Therefore,

a = 1, d = 2, S_{15} = 225

.
Hence, the correct option is 4.

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