Question

Find 11th term from last term of the arithmetic progression10,7,4,...,(-62).​

Answer 1

$a = 10$

$a_n = (-62)$

$d= a_2-a_1$

  $= 7-10\\=(-3)$

$a_1_1 = a+10d$

      $= 10+(10)(-3)\\= 10-30\\=(-20)$

$a_n = (-62 )$

as it is given in question

Answer 2

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.

'a' = 10, 'd' = -3, last term = -62

tn = a + (n − 1)d

⟹ −62 = 10 + (n − 1)( − 3)

⟹ −62 = 13 − 3n

⟹ 3n = 75 or 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.

So,

t15 = 10 + (15 − 1)( − 3)

⟹ t15 = 10 + (14)(− 3)

⟹ t15 = −32