Question

7. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

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Answer 1

a11 = 38

a11 = 38a16 =73

a11 = 38a16 =73 a+15d=73

a11 = 38a16 =73 a+15d=73 - a+10d=38

a11 = 38a16 =73 a+15d=73 - a+10d=38 ▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️

a11 = 38a16 =73 a+15d=73 - a+10d=38 ▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️ 5d = 35

a11 = 38a16 =73 a+15d=73 - a+10d=38 ▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️ 5d = 35 d =7

a11 = 38a16 =73 a+15d=73 - a+10d=38 ▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️ 5d = 35 d =7a+10d=38

a11 = 38a16 =73 a+15d=73 - a+10d=38 ▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️ 5d = 35 d =7a+10d=38a+70 =38

a11 = 38a16 =73 a+15d=73 - a+10d=38 ▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️ 5d = 35 d =7a+10d=38a+70 =38a=38-70

a11 = 38a16 =73 a+15d=73 - a+10d=38 ▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️▪️ 5d = 35 d =7a+10d=38a+70 =38a=38-70 = -32

a31=a+30d

= -32+30x7

= -32+210

=178

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Answer 2

Answer :

$\large\star\:\:\boxed{\dag \:a_{31}=178}\:\:\star$

Explanation :

$\dag$ Given :–

• a₁₁=38
• a₁₆=73

$\dag$ To Find :–

• a₃₁ (The 31st Term of this A.P.)

$\dag$ Formula Applied :–

• $a_n=a+[(n-1)\:\times\:d]$

$\dag$ Solution :–

We have,

=> a₁₁ =38

→ a₁₁ = a + [(11-1) × d]

→ 38 = a + 10d ----------------(1)

=> a₁₆ = 73

→ a₁₆ = a + [(16-1) × d]

→ 73 = a + 15d ----------------(2)

Subtracting Equation(2) from Equation(1) :-

⇒ 73 - (38) = a +15d - (a + 10d)

⇒ 73 - 38 = a - a + 15d - 10d

⇒ 35 = 5 × d

$\implies d=\dfrac{35}{5}$

$\bigstar\:\:\boxed{d=7}$

Now, Putting the value of d in equation(1) :-

⇒ 38 = a + (10 × 7)

⇒ 38 = a + 70

$\implies a=38-70$

$\bigstar\:\:\boxed{a=(-32)}$

So, we have now : a = (-32) , d = 7 , n = 31.

We know the formula : $a_n=a+[(n-1)\:\times\:d]$

Putting the above values in the Formula :-

➼ a₃₁ = (-32) + [(31-1) × (7)]

➼ a₃₁ = (-32) + [(30) × (7)]

➼ a₃₁ = 210 - 32

➼ a₃₁ = 178

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Additional Information :–

A.P.(Arithmetic Progression) is a sequence of series where   after each term of the series, there is a Common Difference (d).

                                                OR

An Arithmetic Progression is a sequence in which the difference between any two consecutive terms is constant which is always denoted by d.

Sum of n terms is given by :

$S_n=\frac{n}{2} [2a+(n-1)d]$

                OR

$S_n=\frac{n}{2} (a+l)\:\:\:\:\:\:\:\:\:[where\:l\:is\:the\:last\:term.]$

Common Difference is given by :

$d=a_{n+1}-a_n$