Answer:

Here, The terms of given A. P. are=(-4),2,8,14,20,26............and so on.

Step-by-step explanation:

Here, As per our given question,

=Sum of the first 15 terms of given A. P. =570

=S15=570

=S15=n/2[2a+(n-1)×d (Where n=15)

=570=15/2[2a+(15-1)×d

=570=15/2(2a+14d)

=570=(15/2×2a)+(15/2×14d)

=15a+105d=570

=a+7d=38 -(1st)eq. (As all numbers are divisible by 15)

Now, As given in question,

=Twelfth term of this A. P.=62

=T12=a+(n-1)×d (Where n=12)

=62=a+(12-)×d

=62=a+11d

=a+11d=62 -(2nd)eq.

Now, As we would apply ELIMINATION method to both equations, Here both equations have same sign (+), So, here we would subtract EQUATION 2 from 1,

= a+7d=38

=-(a+11d=62)

After subtracting, we get,

=7d-11d=38-62

=(-4d)=(-24)

=4d=24 (As minus is on both sides, so it gets cancelled)

=d=24/4

=d=6

Now, by putting the value of d in eq. 1,we get,

=a+7×(6)=38

=a+42=38

=a=38-42

=a=(-4)

So, Now, As we have,

First term(a) of the A. P.=(-4)

Common difference(d) of the A. P.=6

So, Now its sequence or its terms are,

=a, (a+d), a+2d, a+3d, a+4d, a+5d...........

.. (And continue it like this for more terms)

=(-4),(-4+6),(-4+2×6),(-4+3×6),(-4+4×6),(-4+5×6)......

=(-4),2,8,14,20,26............ (Answer).

Thank you.