Question

How many terms are there in the A.P. whose first and fifth terms are 14 and 2 respectively and the sum of the terms is 40?

Answer 1

Answer:

The number of terms of Arithmetic progression is 5 .

Step-by-step explanation:

Given as :

For Arithmetic progression

first term of A.P = 14

fifth term of A.P = 2

Let The number of terms = n

So, for A.P ,

nth terms is written as , $t_n$ = a + [ ( n - 1 ) d ]

where a is first term

And d is common difference

for first term , n = 1

Or, $t_1$ = a + [ ( 1 - 1 ) d ]

Or, 14 = a + 0

i.e  a = 14                     ............1

Again

for fifth term , n = 5

Or, $t_5$ = a + [ ( 5 - 1 ) d ]

Or, 2 = a + 4 d

i.e  a + 4 d = 2

Put the value of a

So, 14 + 4 d = 2

Or, 4 d = 2 - 14

or, 4 d = - 12

∴   d = $\dfrac{-12}{4}$

i.e  d = - 3              ...........2

So, The first term of A.P = a = 14

And The common difference = d = - 3

Again

The sum of n terms of A.P = 40

∵  For A.P

$S_n$ = $\dfrac{n}{2}$ [ 2 a + ( n - 1 ) d ]

put the value of a and d from eq 1 & 2

Or, 40 =  $\dfrac{n}{2}$ [ 2 (14) + ( n - 1 ) (-3) ]

Or, 40 = $\dfrac{n}{2}$ [ 28 - 3 n + 3 ]

Or, 80 = n ( 31 - 3 n )

Or, - 3 n² + 31 n - 80 = 0

i,e  3 n² - 31 n + 80 = 0

Solving the quadratic equation

So, n = $\dfrac{31\pm \sqrt{(-31)^{2}-4\times 3\times 80}}{2\times 3}$

Or, n = $\dfrac{31\pm \sqrt{961-960}}{6}$

Or, n = $\dfrac{32}{6}$ , $\dfrac{30}{6}$

∴   n = 5.3   ,  5

So, The number of terms of A.P = n = 5

Hence, The number of terms of Arithmetic progression is 5 . Answer