Question

find the 31st term of an ap whose 11th term is 880 and 16 term is 73​

Answer 1

Question :

Find the 31ˢᵗ term of an ap whose 11ᵗʰ term is 880 and 16ᵗʰ term is 730.

Answer :

• The 31ˢᵗ term of the AP is 280.

Explanation :

Given :

• 11ᵗʰ term of the AP, a₁₁ = 880
• 16ᵗʰ term of the AP, a₁₆ = 73

To find :

• 31ˢᵗ term of the AP, a₃₁ = ?

Knowledge required :

Formula for nᵗʰ term of an AP :

⠀⠀⠀⠀⠀⠀⠀⠀⠀tₙ = a₁ + (n - 1)d⠀

Where,

• tₙ = nᵗʰ term of the AP.
• a₁ = First term of the AP.
• n = No. of terms of the AP.
• d = Common Difference of the AP.

Solution :

First let us find the nᵗʰ term for the 11ᵗʰ term of the AP :

By using the formula for nᵗʰ term of an AP and substituting the values in it, we get :

⠀⠀=> tₙ = a₁ + (n - 1)d

⠀⠀=> 880 = a₁ + (11 - 1)d

⠀⠀=> 880 = a₁ + 10d

⠀⠀⠀⠀⠀⠀∴ 880 = a₁ + 10d ⠀⠀⠀⠀⠀⠀....(i)

Hence the nth term for the 11ᵗʰ term is [a₁ + 10d]

Now let us find the nᵗʰ term for the 16ᵗʰ term of the AP :

By using the formula for nᵗʰ term of an AP and substituting the values in it, we get :

⠀⠀=> tₙ = a₁ + (n - 1)d

⠀⠀=> 730 = a₁ + (16 - 1)d

⠀⠀=> 730 = a₁ + 15d

⠀⠀⠀⠀⠀⠀∴ 740 = a₁ + 15d ⠀⠀⠀⠀⠀⠀....(ii)

Hence the nᵗʰ term for the 16ᵗʰ term is [a₁ + 15d].

Now,

By subtracting Eq.(ii) from Eq.(i), we get :

⠀⠀=> t₁₁ - t₁₆

⠀⠀=> 880 - 730 = (a₁ + 10d) - (a₁ + 15d)

⠀⠀=> 150 = a₁ + 10d - a₁ - 15d

⠀⠀=> 150 = 10d - 15d

⠀⠀=> 150 = -5d

⠀⠀=> 150/(-5) = d

⠀⠀=> (-30) = d

⠀⠀⠀⠀⠀⠀∴ d = -30

Hence the common difference of the AP (-30).

By substituting the value of d in the equation.(ii), we get :

⠀⠀=> 730 = a₁ + 15(-30)

⠀⠀=> 730 = a₁ - 450

⠀⠀=> 730 + 450 = a₁

⠀⠀=> 1180 = a₁

⠀⠀⠀⠀⠀⠀∴ a = 1180

Hence the First term of the AP 1180.

Now, to find the 31ˢᵗ term of the AP :

By using the formula for nᵗʰ term of an AP and substituting the values in it, we get :

⠀⠀=> tₙ = a₁ + (n - 1)d

⠀⠀=> t₃₁ = 1180 + (31 - 1)(-30)

⠀⠀=> t₃₁ = 1180 + 30(-30)

⠀⠀=> t₃₁ = 1180 + 900

⠀⠀=> t₃₁ = 280

⠀⠀⠀⠀⠀⠀∴ t₃₁ = 280

Therefore,

• 31ˢᵗ term of the AP, a₃₁ = 280.

Answer 2

Solution

We know that

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

Given 11th term is 38

$\sf a_{11} = a + (11 - 1)d \\ \\ \sf \: 38 = a + (11 - 1)d \\ \\ \sf \: 38 = a + 10 \: d \\ \\ \sf \: 38 - 10 \: = a \\ \\ \sf \: a = 38 - 10 \: d........(1)$

Given 16th term is 73

$\sf a_{16} = a + (16 - 1)d \\ \\ \sf \: a_{16} = a + 15d \\ \\ \sf73 = a + 15d \\ \\ \sf \: 73 - 15d = a \\ \\ \sf \: a = 73 - 15d.....(2)$

From (1) and (2)

38 - 10 d = 73 - 15 s

38 - 73 = - 15d - 10d

-35 = -5d

$\sf \dfrac{ - 35}{ - 5} = d$

7 = d

d = 7

Putting value of d in ( 1 )

a = 38 - 10d

a = 38 - 10 × 7

a = 38 - 70

a = - 32

We need to find the 31st term

so,

• n = 31

• a = 32

• d = 7

$\sf \: we \: need \: to \: find \: a_{n}$

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

Putting values

$\sf a_{31} = - 32 \: (31 - 1)7 \\ \\ \sf = \: - 32 + 30 \times 7 \\ \\ \sf = - 32 + 210 \\ \\ \sf \: = 178$

Hence, The 31st term of AP is 178