Question

the 12th term of an AP whose first two terms are -3 and 4 is....​

Answer 1

Answer:

1661

Step-by-step explanation:

Let first term of AP = a

Last term of AP = l

common difference = d

number of terms in AP = n

nth term in AP = a + (n-1)d

So 22th term = a + (22-1)d = a + 21*d (Given n =22)

Given that

22th term = 149

=> a + 21*d = 149

=> a + 21*7 = 149 (Given d = 7)

=> a + 147 = 149

=> a = 149 - 142

=> a = 2

Now sum of AP = (n/2)*{2a + (n-1)*d}

=> (22/2)*{2*2 + (22-1)*7}

=> 11*(4 + 21*7)

=> 11*(4 + 147)

=> 11*151

=> 1661

So first terms in AP is 2 and sum is 1661

I HOPE IT HELPS U

Answer 2

Step-by-step explanation:

a12=?

a=-3

d= a2-a1

d=4-(-3)

d=4+3

d=7

a12= a +11d. ( using general formula)

a12=-3+11(7)

a12= -3+77

a12= 74