Question

the nth term of an AP is given by tn=4n-5. find the sum of first 25 term of the AP​

Answer 1

Step-by-step explanation:

here ,

1st term = 4×1 - 5

= -1

25th term = 4×25 - 5

= 96

now , summation upto 25th term

= 25/2 { ( -1) + 96 }

=25/2 ×95

= 2375/2

Answer 2

Answer:

Sum to 25 terms of the AP = 1175

Step-by-step explanation:

Given,

The nth term of an AP,$t_n$ = 4n -5

To find,

The sum of the first 25 terms of the AP,$S_{25}$

Recall the formula

The sum to n terms of an AP = $S_n = \frac{n}{2}[2a+(n-1)d]$, where 'a' is the first term and 'd' is the common difference

Solution

We have

$t_n$ = 4n -5

$t_1$ = 4×1 -5 = -1

$t_2$ = 4×2 -5 = 3

$t_3$ = 4×3-5 = 7

Hence we have first term a = -1

and common difference, d = $t_2 -t_1$ = 3-(-1) = 4

we have

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

Substitute the value of a= -1,d= 4 and n  = 25

$S_{25} = \frac{25}{2}[2 (-1)+(25-1)4]$

$S_{25} = \frac{25}{2}[ -2+96]$

$S_{25}$ = 25×47 = 1175

∴Sum to 25 terms of the AP = 1175

#SPJ3