an AP consists of 50 terms of which 3rd term is 12 and last term is 106 . find the 29th term or find the sum of all multiples of 7 lying between 500 and 900
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firstly do ,
a+49d= 106
a+2d=12
then ,
the value of a and d will come
so ,
the value of a and b will put in
a +28d so the answer will come of 29 th term
second ,
put the number will come to divisible by 7 between 500 to 900
first number will a and second will be then
minus the second to first
so,
the a and d will come
putt. the formula sn= n/2(2a+(n-1)d
so , the sum will come
I HOPE THIS HELPS U
a+49d= 106
a+2d=12
then ,
the value of a and d will come
so ,
the value of a and b will put in
a +28d so the answer will come of 29 th term
second ,
put the number will come to divisible by 7 between 500 to 900
first number will a and second will be then
minus the second to first
so,
the a and d will come
putt. the formula sn= n/2(2a+(n-1)d
so , the sum will come
I HOPE THIS HELPS U
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