Q. 74.3( 296 Votes )

# Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73

Answer :

To Find : 31st term.

Given: 11th term of AP, a_{11} = 38 and 16th term of AP, a_{16} = 73

We know that a_{n} = a + (n – 1)d

_{n }= nth terms of AP

a = first term of AP

n = number of terms

d = common difference

Hence,

a_{11 }= a + (11 - 1)d

a_{11} = a + 10d = 38 .......eq(i)

And,

a_{16} = a + (16 - 1)d

a_{16} = a + 15d = 73 ........eq(ii)

Subtracting eq(i) from eq(ii), we get following:

a + 15d – (a + 10d) = 73 – 38

a + 15d - a - 10 d = 35Or, 5d = 35

Or, d = 7

Substituting the value of d in eq(i) we get;

a + 10 x 7 = 38

Or, a + 70 = 38

Or, a = 38 – 70 = - 32

Now 31st term can be calculated as follows:

a_{31}= a + (31 - 1)d

a_{31} = a + 30d

= - 32 + 30 x 7

= - 32 + 210 = 178

So, 31st term is 178

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