Question

Find the 31th term of an AP whose 11th term is 38 and the 16th term is 73

Answer 1

Answer:

Step-by-step explanation:

let first term of Ap is a and common difference is d

use formula

tn=a+(n-1) d

now

t11=a+(11-1) d=a+10d=38

same way

t16=a+15d=73

solve this equation then

a=-32 and d=7

now t31=a+30d=-32+210=178

Answer 2

Given:-

• 11th term is 38 and the 16th term is 73.

To find:-

• Find the 31th term of an AP...?

Solutions:-

• The 11th term of Ap is 38
• The 16th term of Ap is 73

we know that;

The 11th term of Ap is 38

=> an = a + (n - 1)d

=> a11 = a + (11 - 1)d

=> 38 = a + 10d .............(i).

The 16th term of Ap is 73

=> an = a + (n - 1)d

=> a16 = a + (16 - 1)d

=> 73 = a + 15d .............(ii).

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

${a} \: + \: {15d} \: \: = \: \: {73} \\ {a} \: + \: {10d} \: \: = \: \: {38} \\ \underline{ - \: \: \: \: \: \: \: \: - \: \: \: \: \: \: \: \: = \: \: \: \: \: \: - \: \: \: \: \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: {5d} \: \: \: \: \: \: \: \: = \: \: \: {35}$

=> d = 35/5

=> d = 7

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

=> a + 10d = 38

=> a + 10(7) = 38

=> a + 70 = 38

=> a = 38 - 70

=> a = - 32

So,

=> a31 = a + (31 - 1)d

=> a31 = a + 30d

=> a31 = - 32 + 30 × 7

=> a31 = - 32 + 210

=> a31 = 178

Hence the 31th term of Ap is 178.