Question

The sum of the third and the seventh terms of an A. P is 6 and their product is 8 . find the sum of the first 16 terms of A. P. ​

Answer 1

Answer:

When a=1 & d=  1/2 , S16 = 76

​When a=5 & d= −1/2, S16 = 20

Thus, the sum of the first 16 terms of A. P is 76 or 20

Step-by-step explanation:

a3 + a7 = 6

a+2d + a+6d = 6

2a+8d=6

a+4d=3

∴ a=3-4d ---eqn1

Now a3 × a7 = 8

∴ (a+2d)(a+6d) = 8

∴ (3-4d+2d)(3-4d+6d) = 8

(3-2d)(3+2d) = 8

9-4d²=8

4d²=9-8=1

d²=1/4

∴ d=±1/2

Now, a=3-4d (from eqn1)

Taking d=1/2

∴ a=3-2=1

Hence, AP is 1, 1.5, 2, 2.5 ....

S16=16/2[2×1+ 15×1/2]=8[2+7.5]= 76

Taking d= -1/2

∴ a=3+2=5

Hence, AP is 5, 4.5, 4, 3.5 ....

S16=16/2[2×5 + 15×(-1/2)] = 8[10 - 7.5] = 20

Answer 2

✰ Let:

• Third term = a₃
• Seventh term = a₇

❖ We know that;

aₙ = a + (n - 1)d

♠ Similarly,

a₃ = a + (3 - 1)d

⇒ a₃ = a + 2d    ------ [Equation ①]

a₇ = a + (7 - 1)d

⇒ a₇ = a + 6d   ------ [Equation ②]

According to the question,

♢ Sum of third and Seventh term is 6

❀ a₃ + a₇ = 6

➞ (a + 2d) + (a + 6d) = 6

➞ a + a + 2d + 6d = 6

➞ 2a + 8d = 6

Taking 2 as a common factor,

➝ 2(a + 4d) =  6

➝ a + 4d = 6 ÷ 2

➝ a + 4d = 3   ----- [Equation ③]

☃ Also, it is given that;

☞ a₃ × a₇ = 8

➝ (a + 2d)(a + 6d) = 8       ⇔   {From Equation 1 and 2}

From Equation 3,

➝ (3 - 4d + 2d)(3 - 4d + 6d) = 8

➝ (3 - 2d)(3 + 2d) = 8

➝ 3² - (2d)² = 8

➝ 9 - 4d² = 8

➝ 4d² = 9 - 8

➝ 4d² = 1

➝ d² = 1/4

➝ d = ± 1/2

∴ d = ± 0.5

Case 1: When d = +0.5,

Now, substitute the value of d in Equation 3 to get the value of a

a + 4d = 3

➞ a + 4(0.5) = 3

➞ a + 2 = 3

➞ a = 3 - 2

∴ a = 1

Case 2: When d = -0.5,

a + 4d = 3

➞ a + 4(-0.5) = 3

➝ a - 2 = 3

∴ a = 5

For finding the sum,

• When a = 1 and d = 0.5;

$\bf{S_{n} = \dfrac{n}{2} \times (2a + (n-1)d)}\\\\\\\Rightarrow \sf{S_{n}= \dfrac{16}{2} \times (2+(16-1)0.5)}\\\\\\\Rightarrow \sf{S_{16} = 8 \times (2 + (15 \times 0.5))}\\\\\\\Rightarrow \sf{S_{16} = 8 \times (2 + 7.5)}\\\\\\\Rightarrow \sf{S_{16} = 8 \times 9.5}\\\\\\\therefore \boxed{\boxed{\bf{S_{16} = 76}}}$

• When a = 5 and d = -0.5

$\bf{S_{n} = \dfrac{n}{2} \times (2a+(n-1)d)}\\\\\\\Rightarrow \sf{S_{16} = \dfrac{16}{2} \times (10 + (16-1)(-0.5))}\\\\\\\Rightarrow \sf{S_{16} = 8 \times (10+[(15)(-0.5)])}\\\\\\\Rightarrow \sf{S_{16} = 8 \times (10-7.5)}\\\\\\\Rightarrow \sf{S_{16} = 8 \times 2.5}\\\\\\\therefore \boxed{\boxed{\bf{S_{16} = 20}}}$

∴ Sum of 16 terms is 76 or 20