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

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Solution

$S_{9} = 81$ $=> \frac{9}{2} [2 a + (9 - 1) d] = 81$
$S_{20} = 400$ $=> \frac{20}{2} [2 a + (20 - 1) d] = 400$
$=> 18 a + 72 d = 162$
$=> 20 a + 190 d = 400$
$=> 2 a + 19 d = 40$
$18 a + 72 d = 162 18 a + 171 d = 360______________- 99 d = - 198 d = 2.$

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Q. The sum of the first 9 terms of an AP is 81 and that of its first 20 terms is 400. Find the first term and the common difference of the AP. [CBSE 2009]

The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of the first 2n terms of another AP, whose first term is - 30 and the common difference is 8. find n?

Q. The sum of the first

9

terms of an

A . P

81

and the sum of it's first

20

terms is

400.

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15 t h

term.

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n

terms of an

A P

whose first term is

8

and the common difference is

20

is equal to the sum of first

2 n

terms of another

A P

whose first term is

40

and the common difference is

6

. Find

n

Q. Question 33
The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is –30 and the common difference is 8. Find n.

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The sum of first 9 terms of an AP is 81 and the sum of its first 20 terms is 400. Find the first term and the common difference of the AP.

S9=81 =>92[2a+(9−1)d]=81S20=400 =>202[2a+(20−1)d]=400=>18a+72d=162=>20a+190d=400=>2a+19d=4018a+72d=16218a+171d=360______________−99d=−198d=2.

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

$S_{9} = 81$ $=> \frac{9}{2} [2 a + (9 - 1) d] = 81$
$S_{20} = 400$ $=> \frac{20}{2} [2 a + (20 - 1) d] = 400$
$=> 18 a + 72 d = 162$
$=> 20 a + 190 d = 400$
$=> 2 a + 19 d = 40$
$18 a + 72 d = 162 18 a + 171 d = 360______________- 99 d = - 198 d = 2.$