Question

three numbers are in AP. if second term is 4,then find the sum of three terms.​

Answer 1

Step-by-step explanation:

Let three yerms be x, 4,4+(4-x)

Sn=n[t¹+tn]

2

=3 [x+(16-x)]

2

=

Answer 2

Answer:

The sum of three terms is 12

Step-by-step explanation:

Given that the three numbers are in AP.

Also given that the second term is 4

The sequence of AP can be written as $a_1,a_2,a_3,....$

Therefore $a_2=4$ ( by given )

Let x be the first term of the sequence

$a_1=x$

Since given numbers are in AP we have common difference d=$a_2-a_1$

$d=4-x$

General formula for AP is $a_n=a+(n-1)d$

• Put n=1 in the above equation we have

$a_1=x+(1-1)(4-x)$

$a_1=x+0(4-x)$

Therefore $a_1=x$

• Put n=2 in the equation $a_n=a+(n-1)d$ we have

$a_2=x+(2-1)(4-x)$

$a_2=x+1(4-x)$

$a_2=x+1(4)+1(-x)$

$=x+4-x$

Therefore $a_2=4$

• Put n=3 in $a_n=a+(n-1)d$

$a_3=x+(3-1)(4-x)$

$=x+2(4-x)$

$=x+2(4)+2(-x)$

$=x+8-2x$

$=8-x$

Therefore $a_3=8-x$

Therefore the sequence $a_1,a_2,a_3,...$ becomes $x,4,8-x,...$

Now sum the first three terms of the AP

x+4+8-x=12

The sum of three terms is 12