Question

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Answer 1

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Let a and d be the first term and the common difference of the given AP respectively. Then,

a

19

=52⇒a+18d=52 ...(1)

a

38

=148⇒a+37d=148 ...(2)

On subtracting (1) from (2), we get

19d=96⇒d=

19

96

Putting d=

19

96

in (1), we get

a=52−18×

19

96

=

19

−740

Now, S

56

=

2

56

[2a+(56−1)d]

∴S

56

=28[

19

3800

]=5600

Answer 2

Given terms are in AP

and  A(19th) + A(38th) = 56

In AP

A(n th term) = a + (n-1)d

        A(19th)= a+(19-1)d = a + 18d........(1)

        A(38th)= a+(38-1)d = a + 37d......(2)

       

Adding (1) and (2)

                A(19th) + A(38th) = 56 = a + 18d + a + 37d

                                      2a+ 55d = 56.......................(3)

                       

Now sum of n terms in AP = S(nth) = $\frac{n}{2} [2a + (n-1)d]$

                                     S(56 th) =    28[ 2a +(56-1)d]

                                                  =  28[ 2a + 55d] = 28(56).....from (3)

                                                  = 1568

The sum of the first 56 terms is 1568

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