Question

Question 48
This is from arthemetic progression

Answer 1

SOLUTION

Here in this A.P., the first term is -4/3 &

Here in this A.P., the first term is -4/3 &last term is 13/3.

The nth term of an A.P. is given by

The nth term of an A.P. is given by a+(n-1)d= Tn,

Here a is the first term,

d is the common difference.

Tn is the last term.

& d= tn+1-tn, here n is natural number.

So,

$d = - 1 - (\frac{ - 4}{3}) \\ = > \frac{ - 3 + 4}{3} = \frac{1}{3}$

So using formula, we get

$= > \frac{ - 4}{3} + (n - 1) \frac{1}{3} = \frac{13}{3} \\ = > \frac{n}{3} = \frac{18}{3} \\ = > n = 18$

So there are 18 term in this AP.

So the two middle most term are 9th and 10th term.

Hence,

$= > t9th = \frac{ - 4}{3} + (9 - 1) \frac{1}{3} = \frac{4}{3} \\ \\ = > t10th = \frac{ - 4}{3} + (10 - 1) \frac{1}{ 3} = \frac{5}{3}$

Hence the sum of two middle most term of the A.P. will be

$= > \frac{4}{3} + \frac{5}{3} \\ \\ = > \frac{9}{3} = 3.$

hope it helps ✔️

Answer 2

Answer:

This is from arthemetic progression

A.P

is the best example for business

t1 = ? , t2 = ?

is the formula of your question

Step-by-step explanation: