Find four consecutive term in an ap whose sum is 12 and sum of third and fourth term is 14 terms are a-d, a, a + d .a+2d
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a-d+a+a+d+a+2d=12
4a+2d=12
a+d+a+2d=14
(2a+3d=14)*2
4a+6d=28
4a+6d-4a-2d= 28-12
4d=16
d=4
4a+6d=28
4a+24=28
4a=4
a=1
a-d=1-4=-3
a=1
a+d=1+4=5
a+2d= 1+8= 9
4a+2d=12
a+d+a+2d=14
(2a+3d=14)*2
4a+6d=28
4a+6d-4a-2d= 28-12
4d=16
d=4
4a+6d=28
4a+24=28
4a=4
a=1
a-d=1-4=-3
a=1
a+d=1+4=5
a+2d= 1+8= 9
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