Question

The second part of q 16

Answer 1

ATQ :

7(a7) = 11(a11)

=> 7 (a + 6d) = 11 ( a + 10d)

=> 7a + 42d = 11a +110d

=> -4a = 68d

=> a = -17d

=> -a = 17d -----------(1)

a18 = a + 17d ( from 1)

=> a18 = a + (-a)

=> a18 = 0

Answer 2

Answer:

Step-by-step explanation:

Given the sum of first 16 terms of theAP is 112

so, 16/2(2a+15d)=112

    =8(2a+15d)=112

    =(2a+15d)=14------(i)

ALSO sum of next forteen terms is 518

so,sum of its 30 terms is 112+518=630

so, 30/2(2a+29d)=630

   =15(2a+29d)=630

   =(2a+29d)=42-------(ii)Subtracting eqn. (i) from (ii).....

           =14d=616

            =d=44

Now putting the value of d we get a= 323