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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:

(i) 3,5,7,9, …. 201

(ii) 3,8,13, … ,253

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

Solution:

In order the find the {{12}^{th}} term for the end of an A.P. which has n terms, its done by simply finding the \left( \left( n-12 \right)+1 \right){}^{th} of the A.P

And we know, {{n}^{th}} term {{a}_{n}}=a+\left( n-1 \right)d

(i) Given A.P =3,5,7,9, …. 201

Here, a=3 and d=\left( 5-3 \right)=2

Now, find the number of terms when the last term is known i.e, 201

{{a}_{n}}=3+\left( n-1 \right)2=201

3+2-2=201

2n=200

n=100

Hence, the A.P has 100 terms.

So, the {{12}^{th}} term from the end is same as {{\left( 100-12+1 \right)}^{th}} of the A.P which is the {{89}^{th}} term.

⇒ {{a}_{89}}=3+\left( 89-1 \right)2

=3+88\left( 2 \right)

=3+176

=179

Therefore, the {{12}^{th}} term from the end of the A.P is 179.

(ii) Given A.P =3,8,13, … ,253

Here, a=3 and d=\left( 8-3 \right)=5

Now, find the number of terms when the last term is known i.e, 253

{{a}_{n}}=3+\left( n-1 \right)5=253

3+5n-5=253

5n=253+2=255

n=255/5

n=51

Hence, the A.P has 51 terms.

So, the {{12}^{th}} term from the end is same as {{\left( 51-12+1 \right)}^{th}} of the A.P which is the {{40}^{th}} term.

⇒ {{a}_{40}}=3+\left( 40-1 \right)15

=3+39\left( 5 \right)

=3+195

=198

Therefore, the {{12}^{th}} term from the end of the A.P is 198.

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