Question

An ap starts with a positive fraction. If the sum of first 11 terms is 33 the the fourth term is

Answer 1

Answer:

The fourth term is 2 .

Step-by-step explanation:

Given as :

The sum of first 11 terms of an A.P = 33

∵ sum of n terms of A.P = $\dfrac{n}{2}$ [2 a + ( n - 1) d]

Or, $S_11$ =  $\dfrac{11}{2}$  × [2 a + ( 11 - 1) d]

Or, 33 =  $\dfrac{11}{2}$  × [2 a + ( 10) d]

Or, 3 × 2 = 2 a + 10 d

Or, 6 = 2 a + 10 d

Diving both side by 2

i.e 3 = a + 5 d                     .............A

Again

As ever alternate is an integer and sum is positive

So, a + 3 d = 2                  .........B

Now, Solvinf eq A and eq B

(a + 5 d) - (a + 3 d) = 3 - 2

i.e (a - a) + (5 d - 3 d) = 1

Or, 0 + 2 d = 1

∴  d = $\dfrac{1}{2}$  = 0.5

common difference = d = 0.5

Put value of d in eq B

So, a + 3 × $\dfrac{1}{2}$ = 2

Or , a = 2 - $\dfrac{3}{2}$

i.e  a = 0.5

First term = a = 0.5

Now, The fourth term =  $t_4$ = a + ( n - 1) d

∴    $t_4$ = 0.5 + ( 4 - 1) 0.5

Or,  $t_4$ = 0.5 + 1.5

Or,  $t_4$ = 2

So, The fourth term =  $t_4$ = 2

Hence, The fourth term is 2 . Answer