Question

In an AP the sum of 11 term is 44 and that of next 11term is 55 find the first term and common difference

Answer 1

Let the AP be a, a+d, a+2d, ...; ==> First term is a and common difference is d  

ii) Sum of first 11 terms is: (11/2)*(2a + 10d) = 44  

This simplifies to: a + 5d = 4 ------ (1)  

iii) Sum of next 11 terms = 55; so sum of first 22 terms is 44 + 55 = 99.  

==> (22/2)(2a + 21d) = 99  

==> 2a + 21d = 9 ------- (2)  

iv) Solving (1) & (2), we have a = 39/11 and d = 1/11  

Hence, AP is; 39/11, 40/11, 41/11, 42/11, ................

hope it help u

Answer 2

Answer:

The first term and common difference for given data is  $\frac{39}{11}$  and $\frac{1}{11}$.

Step-by-step explanation:

In arithmetic Progression (A.P), the sum of n term is given by,

$S_{n} = \frac{n}{2} (2a+(n - 1)d)$

Given that,

• Sum of first $11$ terms $= 44$   (i.e., sum of all terms from first term to $11th$ term)

• Sum of next 11 terms $= 55$  (i.e., sum of all terms from $12th$ terms to $22nd$ terms)

This implies, the sum of first $22$ terms $= 44 + 55$

                                                              $= 99$.

Let the sequence be: $a$, $a+d$, $a+2d$, $a+3d$ $............$

Here, $a =$ First term and,

         $d =$ common difference.

• On substituting sum of first $11$ terms in above equation,

        $S_{n} = \frac{11}{2} (2a+(11 - 1)d)$

        $44 = \frac{11}{2}(2a+10d)$

  $a+5d = 4$

Let this be equation ($1$).

• On substituting sum of first $22$ terms in above equation,

        $S_{n} = \frac{22}{2} (2a+(22 - 1)d)$

         $99 = \frac{11}{2}(2a+21d)$

$2a+21d = 9$

Let this be equation ($2$).

On solving equations ($1$) and ($2$), we get

• $a = \frac{39}{11}$ and,
• $d = \frac{1}{11}$.

Thus, the first term of A.P series is $\frac{39}{11}$ with common difference, $\frac{1}{11}$.