Question

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

Answer 1

Answer:

first term =39/11

common difference = 1/11

Step-by-step explanation:

kindly see the attachment for explanation.

Answer 2

The first term, a = $\frac{39}{11}$  and the common difference, d =  $\frac{1}{11}$ .

Step-by-step explanation:

We are given that in AP the sum of first 11 terms is 44 and that of the next 11 terms is 55.

Let the first term of AP be 'a'  and the common difference be 'd'.

Now, the sum of n terms of an AP formula is given by;

                   $S_n=\frac{n}{2}[2a+(n-1)d]$

We are given, sum of first 11 terms is 44, that means;

                   $S_1_1=\frac{11}{2}[2a+(11-1)d]$

                   $44=\frac{11}{2}[2a+(11-1)d]$  

                   $\frac{44 \times 2}{11}=2a+(11-1)d$

                       $2a + 10d=8$    -------------- [Equation 1]

           

Also, sum of next 11 terms is given as 55, that means;

                   $S_2_2=S_1_1+55$

                   $S_2_2=44+55=99$

So,   $S_2_2=\frac{22}{2}[2a+(22-1)d]$

       $99=\frac{22}{2}[2a+(22-1)d]$  

        $\frac{99 \times 2}{22}=2a+(22-1)d$

              $2a + 21d=9$  ---------------- [Equation 2]

Now, solving equation 1 and 2, we get;

                        $2a + 10d=8$

                        $2a + 21d=9$

                      -     -           -  

                             -11d  =  -1

                                 d  =  $\frac{1}{11}$

So, putting value of d in equation 1, we get;  a = $4 - 5(\frac{1}{11} )$ = $\frac{39}{11}$

Hence, the first term, a = $\frac{39}{11}$  and the common difference, d =  $\frac{1}{11}$ .