In an AP
(i) Given a12 = 37, d = 3, find a and S12.
(ii) Given a3 = 15, S10 = 125, find d and a10.
(iii) Given d = 5, S9 = 75, find a and a9.
(i) Given that, a12 = 37, d = 3
the nth term in an AP,
an = a+(n −1)d,
⇒ a12 = a+(12−1)3
⇒ 37 = a+33
⇒ a = 4
Now, sum of nth term,
Sn = n/2 (a+an)
Sn = 12/2 (4+37)
= 246
(ii) Given that, a3 = 15, S10 = 125
an = a +(n−1)d,
a3 = a+(3−1)d
15 = a+2d …………….. (i)
Sum of the nth term,
Sn = n/2 [2a+(n-1)d]
S10 = 10/2 [2a+(10-1)d]
125 = 5(2a+9d)
25 = 2a+9d …………….. (ii)
On multiplying equation (i) by (ii), we will get;
30 = 2a+4d ………………. (iii)
By subtracting equation (iii) from (ii), we get,
−5 = 5d
d = −1
From equation (i),
15 = a+2(−1)
15 = a−2
a = 17 = First term
a10 = a+(10−1)d
a10 = 17+(9)(−1)
a10 = 17−9 = 8
(iii) Given that, d = 5, S9 = 75
Sn = n/2 [2a +(n -1)d]
S9 = 9/2 [2a +(9-1)5]
25 = 3(a+20)
25 = 3a+60
3a = 25−60
a = -35/3
As we know, the nth term can be written as;
an = a+(n−1)d
a9 = a+(9−1)(5)
= -35/3+8(5)
= -35/3+40
= (35+120/3) = 85/3