Question

Find the A.P whose second terms is 12 and 7th term exeeds the 4th by 15

Answer 1

Given :

• Second term of an AP = 12
• 7th term exceeds the 4th term by 15.

To find :

• AP?

Solution :

• Second term of AP is 12

➾ a₂ = 12

➾ a + d = 12 ❲ eq (1) ❳

★ According to the Question:

• 7th term exceeds the 4th term by 15.

➾ a₇ - a₄ = 15

➾ a + 6d - (a + 3d) = 15

➾ a + 6d - a - 3d = 15

➾ 3d = 15

➾ d = 15/3

➾ d = 5

Now, Putting value of d in eq (1)

➾ a + 5 = 12

➾ a = 12 - 5

➾ a = 7

Now, we have,

• First term, a = 5
• Common difference, d = 7

So, the required AP is,

➻ a , a + d , a + 2d , a + 3d ,.....

➻ 5 , 5 + 7 , 5 + 2(7) , 5 + 3(7) ,....

➻ 5 , 12 , 17 , 26 ,...so on.

Answer 2

(refer to the attachment)

$\large\red\therefore\boxed{\sf{\red{The\:AP\:is\:7,12,17,22,27,32,37.}}}$

• EXPLORE MORE:-

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common difference of 2.

If the initial term of an arithmetic progression is a and the common difference of successive members is d, then the nth term of the sequence an is given by:

$\large\boxed{\sf{a_{n}=a_{1}+(n-1)d}}$

To Find the sum of a series of arithmetic progression,

$\large\boxed{\sf{S_{n}=\dfrac {n}{2}[2a_{1}+(n-1)d]}}$