Question

if the sum of 4th and 6th terms of an ap is 50 and the sum of 12 th and 16th terms is 140 then value of a and d is ​

Answer 1

GIVEN:

• Sum of 4th and 6th term of an AP = 50
• Sum of 12th and 16th term of an AP = 140

TO FIND:

• What is the common difference and first term ?

SOLUTION:

CASE:- 1)

• $\bf{ {a}_{4} + {a}_{6} = 50}$

➨ a + 3d + a + 5d = 50

➨ 2a + 8d = 50

Take common 2 from both sides

➨ a + 4d = 25

➨ a = 25 –4d...❶

CASE:- 2)

• $\bf{ {a}_{12} + {a}_{16} = 140}$

➨ a + 11d + a + 15d = 140

➨ 2a + 26d = 140

Take common 2 from both sides

➨ a + 13d = 70....❷

Put the value of a from equation 1) in equation 2)

➨ 25 –4d + 13d = 70

➨ 9d = 70 –25

➨ 9d = 45

➨ d = $\sf{\cancel\dfrac{45}{9}}$

❮ d = 5 ❯

Put the value of d in equation 1)

➨ a = 25 –4 $\times$ 5

➨ a = 25 –20

❮ a = 5 ❯

❝ Hence, the first term of AP is 5 and common difference is 5 ❞

______________________

Answer 2

Answer:

a=5,d=5

Step-by-step explanation:

formula : a(n)=a+(n-1)d

given

a(4)+a(6)=50

a(12)+a(16)=140