Question

S10 in an Arithmetic progression if the sum of first 3 terms is 9 and sum of next 3 terms is 27 is​

Answer 1

Answer:

$S_{10}$ =100

Step-by-step explanation:

Given,

The sum first three terms of an AP = 9

The sum of the next three terms of AP = 27

To find,

S_10

Recall the concepts,

The terms of the AP is

a a+d, a+2d, .........

Sum of n terms of an AP, $S_n =\frac{n}{2} [2a+(n-1)d]$ --------------(A)

where 'a' is the first term of the AP and 'd' is the common difference.

Solution:

The first three terms of the AP = a,a+d,a+2d

Given, Sum of first three terms of the AP = 9

So, a+a+d+a+2d = 9

3a+3d = 9

a+d = 3 ------------(1)

The next three terms of the AP = a+3d, a+4d, a+5d

Given, Sum of next three terms of the AP = 27

Then, a+3d+a+4d+a+5d = 27

3a+12d = 27

a+4d = 9 ------------(2)

Solving equations(1) and (2)

(2) - (1) -------> 3d = 6

d = 2

From equation(1),

a+2 = 3

a =1

Substituting a = 1, d = 2 and n =10 in Formula(A) we get,

$S_{10} =\frac{10}{2} [2 X1+(10-1)2]$

= 5[2+9×2}

= 5×20

= 100

∴ $S_{10} = 100$

Answer 2

Answer:  $S_{10}$ = 100

Step-by-step explanation:

Given: sum of first 3 terms = 9

           sum of next 3 terms = 27

let the first term is a and the common difference is d

We know that an AP series is

a, a + d, a + 2d, a + 3d,...........

then sum of first 3 terms will be

a + a +d + a +2d = 9

3(a + d) = 9

a + d = 3 ................(1)

Also,

a + 3d + a + 4d + a + 5d = 27

3( a + 4d) = 27

a + 4d = 9............(2)

from equations (1) and (2)

d = 2 ; a = 1

then, Sum of n terms  = $\frac{n}{2}$ {2a + (n - 1)d}

                      $S_{10}$ =  $\frac{10\\}{2}${2 + 18} = 100

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