Question

5. The sum of the first three terms of an AP is 33. If the product of the first term and
third term is 40, then find AP.
please fast!!​

Answer 1

Step-by-step explanation:

a3=a+2d

s3=33

sum of n terms =n(a+an)/2

sum of 3 terms =3*(a+a3)/2=33

a+a3=22

a+a+2d=22

a+d=11

d=11-a

a*(a+2d)=40

a^2+2ad=40

a^2+2a(11-a)=40

a^2+22a-2a^2=40

a^2-22a+40=0

a^2-20a-2a+40=0

a(a-20)-2(a-20)=0

a=2,a=20

if a=2,d=9

if a=20,d=-9

Ap is 2,11,20 or 20,11,2

Answer 2

Given

We have given sum of first three terms of an AP is 33.

The Product of the first and third term is 40.

To find

We have to find the AP.

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Since,we don't know the first term ,second term or common difference so,we have to let .

Let the first term be 'a'

second term be 'a+d'

and the third term be 'a+2d'

Now, according to the question:

Sum of the first three terms is 33

So, a+a+d+a+2d=33

=>3a+3d=33

=>3(a+d)=33

=>a+d=11

a=11-d----(1)

Also ,given the product of the 1st term and 3rd term is 40

First term =a

third term = a+2d

product = a(a+2d)= a²+2ad

=>a²+2ad=40

From equation 1 put a's value into equation.

=>(11-d)²+2(11-d)d=40

identity used here :

(a-b)²=a²+b²-2ab

=>(11)²+d²-2(11)(d)+2[11d-d²]=40

=>121+d²-22d+22d-2d²=40

=>121+d²-2d²-22d+22d=40

=>121-d²=40

=>-d²=40-121

=>-d²= -81

=>d²=81

=>d=√81=9

d=9

put d's value into equation 1

a=11-d

a=11-9

a=2

Hence,first term(a)=2

and common difference (d)=9

AP = a ,a+d,a+2d

AP=2,2+9,2+2×9

AP=2,11,20