Question

The sum of the 4th term and 10th term of an AP is 68 and the sum of 12th and 19th term of an AP Is 136 then find the first 3 terms of AP

Answer 1

Given :-

$\sf a_4+a_{10}= 68\\ \\ \sf a_{12}+a_{19}=136$

To find :-

The first 3 terms of AP

Now

A T.Q :-

we know that

$\bigstar{\boxed{\sf{a_n=a+(n-1)d}}}$

$\sf\ a_4 =a+(4-1)d\\ \\ \longmapsto \sf a_4= a+3d\\ \\ \sf a_{10}= a+9d\\ \\ \longmapsto\sf 4th \ term +10th \ term =68\\ \\\longmapsto \sf a+3d+a+9d=68\\ \\ \longmapsto\sf 2a+12d=68-----(i)$

$\sf a_{12}= a+11d\\ \\ \sf a_{19}= a+18d\\ \\ \longmapsto\sf 12th\ term +19th \ term =136 \\ \\\longmapsto\sf a+11d+a+18d=136\\ \\ \longmapsto\sf 2a+29d=136 ------(ii)$

• Subtract equation (i) From equation (ii)

$\longmapsto\sf \cancel{2a}+29d-\cancel{2a}+12d=136-68\\ \\ \longmapsto\sf 17d= 68\\ \\ \longmapsto\sf d=\cancel{\dfrac{68}{17}}\\ \\ \longmapsto\sf d=4$

• Now we have the value of common difference (d)

• So find the value of a ,put the value of d in eq. (i)

$\longmapsto \sf 2a+12d=68\\ \\ \longmapsto\sf 2a+12\times 4=68\\ \\ \longmapsto\sf 2a=68-48\\ \\ \longmapsto\sf 2a=20\\ \\ \longmapsto\sf a=\cancel{\dfrac{20}{2}}\\ \\ \longmapsto \sf a=10$

• Now find the First 3 terms of AP

we know that

$\sf a_1= a \\ \\\longmapsto \sf a_1= 10\\ \\\sf a_2= a+d\\ \\\longmapsto \sf a_2= 10+4\\ \\\longmapsto \sf a_2= 14\\ \\ \sf a_3= a+2d\\ \\\longmapsto\sf a_3= 10+2\times 4\\ \\\longmapsto\sf a_3= 10+8\\ \\\longmapsto \sf a_3= 18$

So the first 3 terms of AP are

$\boxed{\sf{10,\ \ 14 \ \ 18}}$

Answer 2

GIVEN:

Sum of 4th term and 10th term of an AP = 68

Sum of 12th term and 19th term of an AP = 136

TO FIND:

Find the first 3 terms of AP

SOLUTION:

4th term: a + 3d ; 10th term = a + 9d : 12th term = a + 11d and 19th term = a + 18d

Sum of 4th term and 10th term of an AP = 68

==> a + 3d + a + 9d = 68

==> 2a + 12d = 68 ---(1)

Sum of 12th term and 19th term of an AP = 136

==> a + 11d + a + 18d = 136

==> 2a + 29d = 136 ---(2)

After subtracting both eq - (1) & (2)

==> 17d = 68

==> d = 68/17

==> d = 4

Common Difference = 4

Substitute (d) in eq - (1)

2a + 12d = 68

2a + 12(4) = 68

2a + 48 = 68

2a = 68 - 48

==> 2a = 20

==> a = 20/2

==> a = 10

First term (a) = 10

Second term = a + d = 10 + 4 = 14

Third term = a + 2d = 10 + 8 = 18

Therefore, the first three terms of AP are 10, 14, 18...