Find a Ap Whose 2ndterm is10and the 6th term exceeds 4th term by 12?

Question

Answer

Solution:

Given: second term = 10
and sixth term - fourth term = 12

Now, second term = a+d

according to question,
sixth term - fouth term = 12
(a+5d) - (a+3d) = 12
a+5d - a-3d = 12
2d = 12
d = 6
On putting d = 6 in second term = a+d
we get, 10 = a+6
a = 10 - 6
a = 4
Therefore, the arithematic progression is 4, 10 , 16 , 22, , ..... .

Answer

Answer

Finding an AP with 2nd term as 10 and 6th term exceeds 4th term by 12

Introduction

To find an Arithmetic Progression (AP) whose second term is 10 and the sixth term exceeds the fourth term by 12, we need to use the formula for the nth term of an AP.

Formula for the nth term of an AP

The formula for the nth term of an AP is given by: a + (n-1)d, where a is the first term, d is the common difference, and n is the term number.

Using the given information

We know that the second term is 10, which means that a + d = 10. We also know that the sixth term exceeds the fourth term by 12, which means that a + 5d - (a + 3d) = 12. Simplifying this equation, we get 2d = 12, which gives us a value of d = 6.

Finding the first term

Now that we know the value of d, we can use the equation a + d = 10 to find the value of a. Substituting d = 6, we get a + 6 = 10, which gives us a value of a = 4.

Verifying the solution

To verify that we have found the correct AP, we can calculate the sixth term and the fourth term. Using the formula for the nth term of an AP, we get the sixth term as a + 5d = 4 + 5(6) = 34, and the fourth term as a + 3d = 4 + 3(6) = 22. We can see that the sixth term exceeds the fourth term by 12, as required.

Conclusion

Therefore, the AP with second term as 10 and the sixth term exceeding the fourth term by 12 is: 4, 10, 16, 22, 28, 34.

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