Question

Determine arthematic progression whose third term is 16and and seventh term exceeds the fifth term by 12

Answer 1

AnswEr:-

Given:-

• 3rd term of an AP is 16.

• Seventeenth term exceeds the fifth term by 12.

To Find:-

• The Arithmetic Progression.

Formula Used:-

$\tt\bullet\: a_n = a+(n-1)d.$

Where,

• $\tt a_n = n^{th}$ term.

• n = Number of Terms.

• a = First Term.

• d = Common Difference.

So ATQ:-

$\tt\mapsto a_3 = 16. \\ \\ \tt\mapsto a+(3-1)d = 16. \\ \\ \tt\mapsto a+2d = 16. ---\:eq(1)$

Also,

$\tt\mapsto a_5 + 12 = a_7. \\ \\ \tt (\because\:7th\: term \: exceeds\:5th\:term\:by\:12) \\ \\ \tt\mapsto [a+(5-1)d]+12 = a+(7-1)d. \\ \\ \tt\mapsto \cancel{a}+4d+12 = \cancel{a}+6d. \\ \\ \tt\mapsto 4d+12=6d. \\ \\ \tt\mapsto 6d-4d = 12. \\ \\ \tt\mapsto 2d = 12. \\ \\ \tt\mapsto d=\cancel\frac{12}{2}. \\ \\ \tt\mapsto d=6.$

So by putting the value of d in eq(1) we get,

$\tt\mapsto a + 2d = 16. \\ \\ \tt\mapsto a + 2(6) = 16. \\ \\ \tt (\because \: d = 6). \\ \\ \tt\mapsto a + 12 = 16. \\ \\ \tt\mapsto a = 16 - 12. \\ \\ \tt\mapsto a = 4.$

So by solving we get,

• a = 4.

• d = 6.

So,

• First term,

$\\ \tt = a = 4.$

• Second Term,

$\\ \tt = a + (2 - 1)d. \\ \\ \tt = a + d . \\ \\ \tt = 4 + 6. \\ \\ \tt = 10.$

• Third Term,

$\\ \tt = a + (3 - 1)d. \\ \\ \tt = a + 2d .\\ \\ \tt = 4 + 2(6). \\ \\ \tt = 4 + 12. \\ \\ \tt = 16.$

Therefore the AP is:-

$\implies$ 4, 10, 16.......

Answer 2

Given ,

The third term of AP is 16 and seventh term exceeds the fifth term by 12

Let , First term and common difference of AP be "a" and "d"

First Condition

$\Rightarrow \sf a + (3 - 1)d = 16 \\ \\ \Rightarrow \sf </p><p>a + 2d = 16 --- (i)$

Second Condition

$\Rightarrow \sf a + (7 - 1)d - ( a + (5 - 1)d ) = 12 \\ \\ \Rightarrow \sf </p><p>a + 6d - a - 4d = 12 \\ \\ \Rightarrow \sf </p><p>2d = 12 \\ \\ \Rightarrow \sf d = \frac{12}{2} \\ \\ \Rightarrow \sf </p><p>d = 6$

Put the value of d = 6 in eq (i) , we get

$\Rightarrow \sf a + 2(6) = 16 \\ \\ \Rightarrow \sf </p><p>a + 12 = 16 \\ \\ \Rightarrow \sf </p><p>a = 16 - 12 \\ \\ \Rightarrow \sf </p><p>a = 4$

As we know that , the general form of an AP is given by

a , a + d , a + 2d , ... , a + (n - 1)d

Thus ,

$\therefore \bold{ \sf \underline{The \: AP \: is \: 4 , 10 , 16 , 22 , .... , 6n - 2}}$