Question

If the 5th term of an AP is 31,and 25th term is 14 more than the 5th term , find the AP

Answer 1

Answer:

a+4d=31 (equation 1)

a+24d=14+(a+4d)

a-a+24d-4d=14

20d=14

d= 14/20

d= 7/10

d= 0.7

putting the value of d in equation 1

i.e (a +4d=31)

a+4×0.7=31

a+2.8=31

a=31-2.8

a=28.2

hence,

AP=28.2, 28.9, 29.6, 30.3, 31

Answer 2

Answer:

Step-by-step explanation:

✧ Given :–

• a₅=31
• a₅+14=a₂₅

✧ To Find :–

• The Sequence of The Arithmetic Progression (A.P.)

✧ Formula Applied :–

• aₙ=a+(n-1)d

✧ Solution :–

   We have,

→ a₅=31

→ a₅=a+(5-1)d

★ 31=a+4d --------------------(1)

→ a₅+14=a₂₅

→ 31+14=a₂₅

→ a₂₅=45

→ a₂₅=a+(25-1)d

★ 45=a+24d -----------------(2)

☆ Now, subtracting Equation(2) From Equation(1) :-

⇒ 45-(31)=a+24d-(a+4d)

⇒ 45-31=a+24d-a-4d

⇒ 14=a-a+24d-4d

⇒ 14=20d

$\implies d=\frac{14}{20}$

$\implies \boxed{d=\frac{7}{10}}$

Putting 'd' in Equation(1) :

$\implies a+4(\frac{7}{10} )=31$

$\implies a+(\frac{14}{5} )=31$

$\implies \frac{5a+14}{5}=31$

$\implies 5a+14=31\times5$

$\implies 5a=155-14$

$\implies 5a=141$

$\implies \boxed{a=\frac{141}{5}}$

Now, we need to find the Sequence of A.P. :-

• $a_1=\frac{141}{5}$

• $a_2=a+d=\frac{141}{5} +\frac{7}{10} =\frac{242+7}{10} =\frac{249}{10}$

• $a_3=a+2d=\frac{141}{5} +2(\frac{7}{10}) =\frac{148}{5}$

• $a_4=a+3d=\frac{141}{5} +3(\frac{7}{10} )=\frac{242+21}{10} =\frac{263}{10}$

$\therefore\: A.P. :- \:\:\frac{141}{5} ,\frac{249}{10} ,\frac{148}{5} , \frac{263}{10}$

✼ More Formulas from A.P. :-

• $S_n=\frac{n}{2} [2a+(n-1)d]$

• $S_n=\frac{n}{2} (a+l)\:\:\:\:\:\:\:\:[l\:is\:the\:Last\:Term.]$

• $d=a_{(n+1)}-a_{(n)}\:\:\:\:\:\:\:\:\:[d\:is\:the\:common\:difference.]$

★ There is one Formula which is used to find any term from the Last term :

• $a_n=l-(n-1)d\:\:\:\:\:\:\:\:\:[l\:is\:the\:Last\:Term]$