Question

The 17th term of an ap is -4 and it's 13th tarm is 20 . Find the ap.

Answer 1

Answer :

The required AP is :

92, 86, 80, 74 ,............

Given :

• The 17th term of an AP is -4
• The 13th term of the AP is 20

To Find :

• The required AP (Arithmetic Progression )

Formula to be used :

If a and d are the first term and common difference of an AP respectively then its nth term is given by :

$\bullet \: \: \bf a_{n}= a + (n - 1)d$

Solution :

Let us consider the first term be a and the common difference be d

Given,

17th term = -4

$\implies\sf {a + (17 -1)d = -4}\\\\ \sf {\implies a + 16d = -4 .......... (1)}$

and again

13th term = 20

$\implies \sf{a + (13-1)d= 20}\\\\ \sf \implies {a + 12d =20 .......... (2)}$

Subtracting (1) from (2) we have :

$\sf{ \implies a + 12d -(a + 16d) = 20 -(-4)}\\\\ \sf{\implies a +12d - a -16d = 20 + 4}\\\\ \sf{\implies -4d =24}\\\\ \sf{\implies d = -\dfrac{24}{4} }\\\\ \bf{\implies d = -6}$

Now using the value of d in (2) we have :

$\sf\implies a + 12 (-6) = 20 \\\\ \sf \implies a - 72 = 20 \\\\ \bf \implies a = 92$

Therefore the first term is 92 and common difference is -6

Thus , the required AP is

= 92 , (92-6), (92 - 12) , (92 - 18) , ...............

= 92 , 86 , 80 , 74 , ...............

Answer 2

☣ GIVEN ☣

The 17th term of an A.P. is -4 and it's 13th term is 20.

• a₁₇ = -4

• a₁₃ = 20

☣ TO FIND ☣

The A.P. whose 17th term is -4 and 13th term is 20.

☣ WE MUST KNOW ☣

aₙ = a + (n - 1)d

Where,

• aₙ is the nth term.

• a(or a₁) is the first term.

• n is the number of terms.

• d is the common difference.

☣ SOLUTION ☣

A/q,

$\sf{a_1_7=a+(17-1)d=-4}\\\\\sf{\longmapsto a+16d=-4}\:\:\:\:\:\:\:\: \cdots (i)$

$\rule{100}2$

$\sf{a_1_3=a+(13-1)d=20}\\\\\sf{\longmapsto a+12d=20}\:\:\:\:\:\:\:\: \cdots (ii)$

Subtracting eq.(ii) from eq.(i) :-

$\sf{a+16d=-4}\\-\\\sf{a+12d=20}\\\rule{60}{0.8}\\\sf{4d=-24}\\\\\\\sf{\longmapsto d=\dfrac{-24}{4} }\\\\\boxed{\sf{\longmapsto d=-6}}$

$\rule{100}2$

Putting the value of d in eq.(i) :-

$\sf{a+16d=-4}\\\\\sf{\longmapsto a+16(-6)=-4}\\\\\sf{\longmapsto a-96=-4}\\\\\sf{\longmapsto a=-4+96}\\\\\boxed{\sf{\longmapsto a=92}}$

Now, an A.P. is of the form :-

a, (a + d), (a + 2d), (a + 3d), ....

Hence the required A.P. is :-

92 , (92 - 6), (92 - 12) , (92 - 18) , ...

= 92 , 86 , 80 , 74 , 68, ...