Question

Example 8 : Find the 11th term from the last term (towards the first term) of the
AP: 10, 7, 4, ..., - 62.​

Answer 1

Answer:

-32

Step-by-step explanation:

The sequence given is 10, 7, 4, ....-62.

First we need to find out number of terms in this sequence.

First term, a = 10

Common difference,d= -3

Last term,l = -62

tn​=a+(n−1)d

⇒−62=10+(n−1)(−3)

⇒−62=13−3n

⇒3n=75

⇒n=25.

The number of terms in the sequence is 25.

The 11th term from the end is same as the 15th term from the first.

$t_{n}$​=a+(n−1)d

So,

   $t_{11}$​=10+(15−1)(−3)

⇒$t_{11}$​=10+(14)(−3)

⇒$t_{11}$=−32.

The 11th term from the end is -32.

Answer 2

Answer:

AP: 10, 7, 4, ....,-62

Step-by-step explanation:

Given:

a=10

d=-3

an th term= -62

n= ?

We know that

$an = a + (n – 1) × d$

$- 62 = 10 + (n - 1) ( - 3) \\ - 62 - 10 = (n - 1)( - 3) \\ - 72 = (n - 1)( - 3) \\ \frac{ - 72}{ - 3} = (n - 1) \\ 24 = (n - 1) \\ 24 + 1 = n \\ 25 = n$

There are 25 terms in the AP.

11th term from the last term is

25-10 = 15 (25-10 because 11th term from the last term. In these type of questions we even have to include the last term.)

Therefore, 11th term from 25th term is 15th term.

$a15 = 10 + (15 - 1)( - 3) \\ a15 = 10 + (14)( - 3) \\ a15 = 10 + ( - 42) \\ a15 = 10 - 42 \\ a15 = - 32$

Therefore the 11th term from the last term is -32.