Question

find the last term of the series 7+14+21+....20 terms​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The last term of the series

7 + 14 + 21 +........... 20 terms

FORMULA TO BE IMPLEMENTED

If in an arithmetic progression

First term = a and

common difference = d

Then the n th term of the Arithmetic progression is

$\sf{t_n = a + (n-1)d}$

CALCULATION

Here the given progression is

7 + 14 + 21 +........... 20 terms

It is an Arithmetic progression

First term = a = 7

Common Difference = d = 14 - 7 = 7

Since the last term i.e 20 th term is to be determined

So n = 20

Hence the last term is

$= \sf{t_{20} = a + (20-1)d}$

$\sf{= a +19d}$

$\sf{= 7 +(19 \times 7)}$

$\sf{= 7 +133}$

$\sf{= 140}$

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Answer 2

Given:

series 7+14+21+.......20 terms

so, First term = a = 1

     common difference = d = 14 - 7 = 7

     number of terms = n = 20

To find:

last term of the series =?

Step-by-step explanation:

Given there are 20 terms in the series.

Now, the formula for $n^{th}$ term of a series = $T_{n}$ = a + (n-1×)d

The last term will be the $20^{th$ terms.

So, $T_{20}$ = 7 + (20 - 1)×7

     $T_{20}$ = 140

Hence, the last term of the series is 140