Question

the 18th term if an AP is 30 more then its 8th term. if the 15th term of the AP is 48,find thr AP​

Answer 1

${\huge{\underline{\underline{\sf{\maltese\; Question:–}}}}}$

Q) The 18th term of an AP is 30 more than it's 8th term . If the 15th term of the AP is 48 , Find the AP .

${\huge{\underline{\underline{\sf{\maltese\:Answer\checkmark}}}}}$

☆ Given :–

• 18th term = 30 + 8th term .
• 15th term = 48 .

☆ To Find :–

• The AP .

☆ Solution :–

• Let the first term be → x
• Let the common difference be → d

According to the Question ;

$\rm \:1 8 {}^{th} \: term = 30 + {8}^{th} \: term$

$\mapsto \rm \: a + (18 - 1)d = 30 + a + (8 - 1)d$

$\mapsto \rm \: \cancel{a} + 17d = 30 + \cancel{a }+ 7d$

$\mapsto \rm \: 17d - 7d = 30$

$\mapsto \rm \cancel{10}d = \cancel{30}$

$\mapsto \underline{ \boxed{ \rm{ \pink{ \: d = 3 \: }}}}$

Now ,

It is given that ,

$\rm \: 15 {}^{th} \: term = 48$

$\mapsto \rm \: a + (15 - 1)d = 48$

$\mapsto \rm \: a + 14d = 48$

Put the value of d

$\mapsto \rm \: a + 14 \times 3 = 48$

$\mapsto \rm \: a + 42 = 48$

$\mapsto \rm \: a = 48 - 42$

$\mapsto \underline{ \boxed{ \pink{ \rm \: {a = 6 \: }}}}$

Now ,

The AP would be

${\tt{a,a+d,a+2d,a+3d,...}}$

${\rightarrow{\tt{6,6+3,6+6,6+9,...}}}$

So ,

${\underline{\boxed{\sf{\purple{A.P.→6,9,12,15,...}}}}}$

_________________________

☆ Know More :–

• A.P. → It is a sequence in which adjacent terms differ with a common difference .

• ${\rm{A_n=a+(n-1)d}}$

• ${\rm{S_n=\dfrac{n}{2}\{2a+(n-1)d\}}}$

• ${\rm{S_n=\dfrac{n}{2}\{a+a_n\}}}$

Where ,

$\pink{ \ddot{ \smile}} \: \rm\: a = first \: term$

$\purple{ \ddot{ \smile}} \: \rm \: d = common \: difference$

$\orange{ \ddot{ \smile}} \: \rm \: a _{n} = {n}^{th} \: term$

$\blue{ \ddot{ \smile}} \: \rm \: n = no. \: of \: terms$

$\red{ \ddot{ \smile}} \: \rm \: s _{n} = sum \: of \: n \: terms \:$