Question

The sum of 10 terms of an A.P. is 125. If is third
term is 15. Find the common difference
​

Answer 1

Answer:

common difference = -1

Step-by-step explanation:

Given :

• The sum of 10 terms of an A.P. is 125
• the third term = 15

To find :

the common difference

Solution :

Let a be the first term and d be the common difference.

Sum of n terms is given by,

$\sf S_n=\dfrac{n}{2}[2a+(n-1)d]$

Sum of 10 terms = 125

$\sf S_{10}=\dfrac{10}{2}[2a+(10-1)d] \\\\ \sf 125=5[2a+9d] \\\\ \sf 125/5=2a+9d \\\\ \sf 25=2a+9d \rightarrow [1]$

nth term of an A.P. is given by,

aₙ = a + (n - 1)d

Third term = 15

a₃ = a + (3 - 1)d

15 = a + 2d ➙ [2]

Multiply equation [2] by 2,

2(15) = 2(a + 2d)

30 = 2a + 4d ➙ [3]

Subtract equation [3] from equation [1],

25 - 30 = 2a + 9d - (2a + 4d)

 -5 = 2a + 9d - 2a - 4d

 -5 = 5d

  5d = -5

   d = -5/5

   d = -1

Hence, common difference = -1