Question

An AP consist of three terms whose sum is 15 and sum of their squares of extrems is 58. Find the first three terms of AP and also find the sum of first 50 terms of an AP​

Answer 1

Answer:

hey frnd here is Ur answer ...

Step-by-step explanation:

let the three terms be a-d,a,a+d

you have given sum is 15

a-d+a+a+d=14

so from here a=15

and u have given sum of squares is 58 so by here (a²-d²)(a²)=58

so now from u will get d and then u will get first three terms also...

thnx

hloo hii..

plzz mark as brainlist ...

hloo hii...

Answer 2

Answer with Step-by-step explanation:

Let three terms of A.P are a-d,a and a+d.

According to question

$a-d+a+a+d=15$

$3a=15$

$a=\frac{15}{3}=5$

$(a-d)^2+(a+d)^2=58$

$a^2+d^2-2ad+a^2+d^2+2ad=58$

Using the property :$(a+b)^2=a^2+b^2+2ab$

$2a^2+2d^2=58$

$2(a^2+d^2)=58$

Substitute the value

$2(5^2+d^2)=58$

$25+d^2=\frac{58}{2}=29$

$d^2=29-25=4$

$d=\sqrt{4}=\pm 2$

Substitute a=5 and d=2

$a_1=a-d=5-2=3$

$a_2=a=5$

$a_3=a+d=5+2=7$

Substitute a=5 and d=-2

$a_1=a-d=5-(-2)=7$

$a_2=a=5$

$a_3=a+d=5+(-2)=3$

Sum of nth terms of A.P

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

Substitute d=2,n=50 and a=5

$S_{50}=\frac{50}{2}(2(5)+(50-1)(2))$

$S_{50}=25(10+49\times 2)=2700$

Substitute d=-2 and a=5

$S_{50}=\frac{50}{2}(2\times 5+49\times (-2))=-2200$

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https://brainly.in/question/14765875:Answered by bradlamar