Question

which term of the ap 3,10,17....will be 84 more than its 11th term​

Answer 1

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

• AP = 3, 10 , 17....

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

• we need to find the term which will be 84 more than 11th term.

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

≫ AP = 3, 10 , 17

• First term (a) = 3
• Common difference (d) = 10 - 3 = 7

↱ an = a + (n - 1)d

$\tt \rightarrowtail \: a_ {11} = a + 10d \\ \\ \tt \rightarrowtail \: a_ {11} = 3 + 10 \times 7 \\ \\ \tt \rightarrowtail \: a_ {11} = 3 + 70 \\ \\ \bf\rightarrowtail \: a_ {11} = 73$

Now , finding the term which is 84 more than it's 11th term

$\tt \dashrightarrow \: a_ n = a_{11} + 84 \\ \\ \tt \dashrightarrow \: a_ n = 73 + 84 \\ \\ \bf\dashrightarrow \: a_ n =15 7$

Now finding the term:-

$\tt \dashrightarrow \: 157 = a + (n - 1)d \\ \\ \tt \dashrightarrow \: 157 = 3 + (n - 1)7 \\ \\ \tt \dashrightarrow \:157 - 3 = (n - 1)7 \\ \\ \tt \dashrightarrow \: \cancel\frac{154}{7} = (n - 1) \\ \\ \tt \dashrightarrow \:22 = n - 1 \\ \\ \tt \dashrightarrow \:n = 23$

So,

≫23 is the term which is 84 more than it's 11th term.

$\rule{200}3$

Answer 2

$\sf\huge{\underline{\underline{Question:-}}}$

which term of the ap 3,10,17....will be 84 more than its 11th term.

$\sf\huge{\underline{\underline{Given:-}}}$

• 1st term (a) = 3
• common difference (d) = 7
• $\sf a_{11}=?$

$\sf\huge{\underline{\underline{Identity\:Used:-}}}$

$\large{\fbox{\red{\underline{\blue{a_n= a+(n-1)d}}}}}$

$\sf{\underline{\underline{\red{Substitute \:all\:values\:in\: formula:-}}}}$

$\sf→ a_n=a+(n-1)d\\\sf→ a_{11}=3+(11-1)7\\\sf→ a_{11}=3+(10)7\\\sf→ a_{11}=3+70\\\sf→ {\fbox{\red{\underline{a_{11}= 73}}}}$

Now,

According to question

• we have to find the term 84 more than 11th term .

$\sf→ a_n= a_{11}+84. ( a_{11}=73)\\\sf→ a_n=73+84 = a_n= 157$

Therefore,

$\sf→ a_n=a+(n-1)d \\\sf→ 157=3+(n-1)7\\\sf→ 157-3=(n-1)7\\\sf→ \frac{154}{7}=n-1\\\sf→ 22=n-1 \\\sf{\fbox{\underline{\blue{n=23}}}}$