Question

How many terms are there in an Arithmetic progression whose first term and 6th term are -12 and 8 respectively and sum of all its terms is 120?
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Answer 1

Solution :

First term, a = - 12

A6 = a + 5d

8 = a + 5d

8 = - 12 + 5d

5d = 8 +12 = 20

d = 20 ÷ 5 = 4

d, Common difference = 4

Now, Sn = 120

120 = n/2 [ 2a + ( n - 1) d]

120*2 = n [ 2( - 12) + ( n - 1 )(4)]

240 = n [ - 24 + 4n - 4]

240 = n( 4n - 28)

240 = 4n( n - 7)

n( n - 7) = 240/4

n^2 - 7n = 60

n^2 - 7n - 60 = 0

n^2 - 12n + 5n - 60 = 0

n( n - 12 ) + 5 ( n - 12 ) = 0

( n - 12) ( n +5 ) = 0

n = 12, -5

Since, number of terms cannot be negative and in fraction.

So, there are 12 number of terms.