Question

In an ap of 27 terms the sum of three middle terms is 117 and the sum of its last terms is 321 then find the sum of next three terms ​

Answer 1

Answer:

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

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

∴ (x1 + 24n) + (x1 + 25n) + (x1 + 26n) = 321

∴ 3x1 + 75n = 321   ...... (3)

∴ 3x1 = 321 - 75n..

substituting the value in equation 2, we get

(321 - 75n) + 39n = 177

∴ 321 - 177 = 75n - 39n

∴ 144 = 36n

∴ n = 4 .... (4)

Substtuting in eqn (3),

3x1 + 75× 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 = 7, x2 = 11, x3 = 15

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