Question

the sum of the third and seventh term of an ap is 32 and their product is 220 find the sum of the first twenty one terms
of an ap​

Answer 1

Solution :-

Let the first term and common difference of an AP be a and d respectively.

Case I : The sum of the third and seventh term of an AP is 32.

=> a + 2d + a + 6d = 32

=> 2a + 8d = 32

=> 2(a + 4d) = 32

=> a + 4d = 16

=> a = 16 - 4d ______(i)

Case II : Their product is 220.

=> (a + 2d) (a + 6d) = 220

=> (16 - 4d + 2d) (16 - 4d + 6d) = 220 [from equation (i)]

=> (16 - 2d) (16 + 2d) = 220

=> (16)² - (2d)² = 220

=> 256 - 4d² = 220

=> 4d² = 256 - 220

=> d² = 36/4

=> d = √9 = 3

Substituting the value of d in equation (i) :

a = 16 - 4 × 3 = 4

Sum of first twenty one term = n/2 [2a + (n - 1)d]

= 21/2 [ 2 × 4 + (21 - 1)3]

= 21/2 [ 8 + 60]

= (21 × 68)/2

= 714

Hence,

Sum of first twenty one term = 714

Answer 2

$\huge\underline\mathfrak\pink{Answer}$

The sum of first twenty one terms of an AP is 714.

_____________________________

$\huge\underline\mathfrak\pink{Explanation}$

We know that,

$a_n$ = a+(n-1)d

where,

a is first term

n is number of terms

d is common difference

★According to question,

a + (3-1)d + a + (7-1)d = 32

a + 2d + a + 6d = 32

2a + 8d = 32

Taking 2 as common factor,

a + 4d = 16

a = 16-4d

___________________________

Also,

(a+2d)(a+6d) = 220 ....(1)

Now, substitute a = 16-4d in eq (1).

We get,

(16-4d+2d)(16-4d+6d) = 220

(16-2d)(16+2d) = 220

(16)² - (2d)² = 220 [ (a+b)(a-b)=a²-b² ]

256 - 4d² = 220

-4d² = 220 - 256

-4d² = -36

4d² = 36

d = ±√(36/4)

d = ±6/2

d = ±3

______________________________

★When d = 3

a = 16 - 4(3)

a = 16 - 12

a = 4

★when d = -3

a = 16-4(-3)

a = 16 + 12

a = 28

___________________________

We also know that,

$S_2_1$ = 21/2[2×4+(20)(3)]

$S_2_1$ = 21/2(8+60)

$S_2_1$ = 21/2(68)

$S_2_1$ = 21×34

$S_2_1$ = 714 ( required answer )