Question

an ap has 27 terms the sum of the middle terms and the two terms adjacent one one each side of it is 177 if sum of the last three terms is 321 find the first three terms of an ap​

Answer 1

Answer:

First three term is AP is 7 , 11 , 16.

Step-by-step explanation:

Given: No of terms in AP, n = 27

          Middle term of AP is 14th term ⇒ Sum of 13th , 14th & 15th term = 177

          Sum of last three term.i.e., 25th , 26th & 27th term = 321

To find: First three terms

Let first term of given AP is a and common difference is d.

using formula of term of an AP, we get

$a_n=a+(n-1)d$

$a_{13}+a_{14}+a_{15}=177$

$a+(13-1)d+a+(14-1)d+a+(15-1)d=177$

$3a+12d+13d+14d=177$

$3a+39d=177$

$a+13d=59$..............................(1)

$a_{25}+a_{26}+a_{27}=321$

$a+(25-1)d+a+(26-1)d+a+(27-1)d=321$

$a+24d+a+25d+a+26d=321$

$3a+75d=321$

$a+25d=107$............................(2)

Using Elimination method in equation (1) & (2) , we get

12d = 48

d = 4

put this value in eqn (1), we get

a + 13 × 4 = 59

a + 52 = 59

a = 59 - 52

a = 7

So, 1st term = 7

    2nd term = 7 + 4 = 11

     3rd term = 11 + 4 = 16

Therefore, First three term is AP is 7 , 11 , 16.

Answer 2

Answer:

here is the best answer

please please please please please mark my answer as brainlist because

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