if a + 1, 3a and 4a plus 2 or 3 consecutive terms of an arithmetic progression find a
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t1 =a+1
t2 =3a
t3=4a+2
t1 , t2 and t3 are term 1 , term 2 ,term 3 respectively
d = t2-t1
d = 3a-a-1
d = 2a-1
d= t3 -t2
2a -1 = 4a+2 -3a
2a -a = 3
a = 3
three terms are
a+1=3+1=4
3 a=3*3=9
4a+2= 4*3+2 =12+2=14
t2 =3a
t3=4a+2
t1 , t2 and t3 are term 1 , term 2 ,term 3 respectively
d = t2-t1
d = 3a-a-1
d = 2a-1
d= t3 -t2
2a -1 = 4a+2 -3a
2a -a = 3
a = 3
three terms are
a+1=3+1=4
3 a=3*3=9
4a+2= 4*3+2 =12+2=14
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