Question

National Standard Examination III JL
79. There are ten numbers in a certain A.P. The sum
of first three terms is 321. The sum of last three
numbers is 405. Find the sum of all the ten
numbers.
(1) 1165
(2) 1210
(3) 1221
(4) 1252
[2015-2016]​

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Answer:

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Step-by-step explanation:

Let the 1st term = x1

for a.p., constant term added = n

.. we can write equation

x2 = x1 + n

x3 = x1 + 2n

xn = x1 + (n-1)n

- (1)

There are total 27 terms

The middle term will be 14th term

.. As per condition (1)

x13 + x14 + x15 = 177

MEa (0) Q 32 32

substituting from equation 1 we get,

x13 = x1 + 12n

x14 = x1 + 13n

x15 = x1 + 14n

:: x13 + x14 + x15 = (x1+12n) + (x1+13n)

+ (x1+14n) = 177

· (2)

::3x1 + 39n = 177.. (2)

For condition 2, sum of last three

terms = 321

x25 + x26 +x27 = 321

MEg (0) Q 32 32

-(X1 + 24n) + (x1 + 25n) + (x1 + 26n) =

321

: 3x1 + 75n = 321

.. 3x1 = 321 - 75n..

(3)

substituting the value in equation 2,

we get

(321 - 75n) + 39n = 177

. 321 - 177 = 75n - 39n

:: 144 = 36n

.. n = 4. (4)

Substituting in eqn (3),

3x1 + 75x 4 = 321

:: 3x1 + 300 = 321

:: 3x1 = 21

.. x1

= 7 . (5)

1st 3 terms of a.p.

x1 = 7

x2 = x1 + n = 7 + 4 = 11

x3 = x2 + 2 = 11+ 4 = 15

: x1

= 7x2 = 11, x 3 = 15