Question

If 7th and 13th terms of an AP be 34 and 64 respectively , then its 18th term is​

Answer 1

Answer:

a+(n-1)d as we know

Step-by-step explanation:

so a+(7-1)d=34

a+6d=34-1st equation

a+(13-1)d=64

a+12d=64-2nd equation

2nd equation-1st equation will give

12d-6d=64-34

6d=30 ,d=5

putting d in any of the equations we will get a=4

now 4+(18-1)5=4+85=89 which is 18th term

Answer 2

Given:-

• The 7th term of Ap = 34
• The 13th term of Ap = 64

To find:-

• Find the 18th term of an Ap.. ?

Solutions:-

• an = a + (n - 1)d

We have,

The 7th term of Ap = 34

=> a7 = a + (7 - 1)d

=> 34 = a + 6d ...........(i).

The 13th term of Ap = 64

=> a13 = a + (13 - 1)d

=> 64 = a + 12d ...........(ii).

Now, Subtracting Eq. (ii) and (i) we get,

${a} \: + \: {12d} \: \: = \: \: {64} \\ {a} \: + \: {6d} \: \: = \: \: {34} \\ \underline{ - \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: = \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: {6d} \: \: \: \: \: \: \: \: = \: \: \: {30}$

=> d = 30/6

=> d = 5

Now, putting the value of y in Eq. (i).

=> a + 6d = 34

=> a + 6 × 5 = 34

=> a + 30 = 34

=> a = 34 - 30

=> a = 4

Now, 18th term of Ap.

=> a18 = a + (18 - 1)d

=> a18 = 4 + 17 × 5

=> a18 = 4 + 85

=> a18 = 89

Hence, the 18th term of Ap is 89.