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13. Find the {{12}^{th}} term from the end of the following arithmetic progressions:
Arithmetic Progression | Class 10 | ICSE | Maths | RD Sharma
13. Find the {{12}^{th}} term from the end of the following arithmetic progressions:

(iii) 1,4,7,10, … ,88

An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

(iii) Given A.P =1,4,7,10, … ,88

Here, a=1 and d=\left( 4-1 \right)=3

Now, find the number of terms when the last term is known i.e, 88

{{a}_{n}}=1+\left( n-1 \right)3=88

1+3n-3=88

3n=90

n=30

Hence, the A.P has 30 terms.

So, the {{12}^{th}} term from the end is same as {{\left( 30-12+1 \right)}^{th}} of the A.P which is the {{19}^{th}} term.

⇒ {{a}_{89}}=1+\left( 19-1 \right)3

=1+18\left( 3 \right)

=1+54

=55

Therefore, the {{12}^{th}} term from the end of the A.P is 55.

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