Question

Find the 31st term of AP whose 11th term is 38 and 16th term is 73 (Question s from arithmetic progressions class 10 cbse) PLEASE ANSWER WITH FULL STEPS

Answer 1

Answer:

hope this helps you....

Answer 2

$\large\bf\underline{Given:-}$

• 11th term = 38
• 16th term = 73

$\large\bf\underline {To \: find:-}$

• 31st term of AP

$\huge\bf\underline{Solution:-}$

we know that,

$\star\: \: \: \underline{ \boxed{ \bf \: T_n = [a+(n-1)d]}}$

11th term = 38

• a + 10d = 38......(i)

16th term = 73

• a + 15d = 73.......(ii)

Solving eq.(i) and (ii)

$\rm \: a + 10d = 38 \\ \rm \: a + 15d = 73 \\ \: - \: \: \: \: \: \: - \: \: \: \: \: - \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \: \: \: \: \: \: \: \: \rm - 5d = - 35 \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: d = 7$

substituting value of d = 7 in eq.(i)

$\longmapsto \rm \: a + 10d = 38 \\ \longmapsto \rm \:a + 10 \times 7 = 38 \\ \longmapsto \rm \:a + 7 0 = 38 \\ \longmapsto \rm \:a = 38 - 70 \\ \longmapsto \bf\:a = - 32$

• So, First term (a) = -32
• So, First term (a) = 7

Now, finding 31st term of AP.

31st term = a + 30d

$\leadsto\rm \:a_{31 } = a+ 30d \\ \leadsto\rm \:a_{31 } = - 32 + 30 \times 7 \\ \leadsto\rm \:a_{31 } = - 32 + 210 \\ \leadsto\bf \:a_{31 } = 178$

So, the 31st term of AP = 178