Question

there are 20 terms in an arithmetic sequence. sum of first and last term is 88. if 10th term 42,what is the 11th term?​

Answer 1

Given :-

◉ There are 20 terms in an AP,

• Sum of first and the last term is 88.
• 10th term of the AP is 42.

To Find :-

◉ 11th term of the AP

Solution :-

It is given in the question that there are 20 terms in the AP, Also the sum of first term and the last term is 88.

But, There are 20 terms in the AP,

∴ First term + 20th term = 88

⇒ a + a + (20 - 1)d = 88

⇒ 2a + 19d = 88 ...(1)

Again, It is given that 10th term of the AP is 42.

∴ a + (10 - 1)d = 42

⇒ a + 9d = 42

Multiply both sides by 2,

⇒ 2a + 18d = 84 ...(2)

Subtract (1) from (2), we get

⇒ 2a + 18d - 2a - 19d = 84 - 88

⇒ -d = -4

⇒ d = 4

Substituting the value of d in (1), we have

⇒ 2a + 19×4 = 88

⇒ 2a = 88 - 76

⇒ 2a = 12

⇒ a = 6

Now, That we have got the value of a and d, we can find the eleventh term,

We know,

⇒ aₙ = a + (n - 1)d

Put n = 11, because we have to find the eleventh term,

⇒ a₁₁ = a + 10d

Substituting the values of a and d,

⇒ a₁₁ = 6 + 10×4

⇒ a₁₁ = 6 + 40

⇒ a₁₁ = 46

Hence, The eleventh term of the AP is 46.

Answer 2

$\mathcal{\gray{\underbrace{\blue{GIVEN:-}}}}$

• There are 20 terms in an arithmetic progress [A.P] .

• The sum of 1st and last [20th term] is 88 .

• 10th term of the A.P is 42 .

$\mathcal{\gray{\underbrace{\blue{TO\: FIND:-}}}}$

• The 11th term of the A.P .

$\mathcal{\gray{\underbrace{\blue{SOLUTION:-}}}}$

$\orange\bigstar\:\mathcal{\pink{\boxed{\purple{a_n\:=\:a\:+\:(n-1)d}}}}$

Where,

• a = first term .

• d = common difference .

• ‘n’ represents no. of term .

⚡ According to the question,

=> first term + last term = 88 .

$\rm{\red{\implies\:a\:+\:a_{20}\:=\:88\:}}$

=> a + [a + (20 - 1) d] = 88

=> a + a + 19d = 88

=> 2a + 19d = 88 -------(1)

Again,

=> 10th term of the A.P = 42

$\rm{\red{\implies\:a_{10}\:=\:42\:}}$

=> [a + (10 - 1) d] = 42

=> a + 9d = 42 --------(2)

✍️ Multiply 2 in equation (2), we get

=> 2a + 18d = 84 --------(3)

✍️ Now substract equ(1) and equ(3), we get

=> (2a + 19d) - (2a + 18d) = 88 - 84

=> 2a + 19d - 2a - 18d = 4

=> d = 4

⚡ Putting the value of ‘d’ in equation (2), we get

=> a + (9 × 4) = 42

=> a + 36 = 42

=> a = 42 - 36

=> a = 6

✍️ Now, the 11th term of the A.P is

$\rm{\green{\implies\:a_{11}\:=\:a\:+\:(11-1)\:d}}$

$\rm{\green{\implies\:a_{11}\:=\:6\:+\:10\times{4}\:}}$

$\rm{\green{\implies\:a_{11}\:=\:6\:+\:40\:}}$

$\rm{\red{\boxed{\green{\implies\:a_{11}\:=\:46\:}}}}$

$\rm{\pink{\therefore}}$ The 11th term of the A.P is “46” .