Answer :

Given, first term a = – 5

Last term t_{n} = 45

Sum of n terms S_{n} = 120

To find no of terms “n”

Using Sum of n terms of an A.P. formula

where n = no. of terms

S_{n} = sum of n terms

Now, on substituting given value in formula we get,

⇒ 120 = 20 n

To find the common difference ‘d’

We use n^{th} term of an A.P. formula

t_{n} = a + (n – 1)d

where n = no. of terms

a = first term

d = common difference

t_{n} = n^{th} terms

Thus, on substituting all values we get,

⇒ t_{6} = – 5 + (6 – 1)d

⇒ 45 = – 5 + 5d

⇒ 5d = 45 + 5 = 50

Thus, common difference is 10

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