Question

The sum of the first 4 terms of an arithmetic progression is 142 .The
11th term is two more than twice the
3rd term
Find the common difference.​

Answer 1

To Find :-

⠀⠀⠀⠀⠀⠀⠀The Common Difference of the AP.

Given :-

• Sum of first 4 terms = 142

We know :-

⠀⠀⠀⠀⠀⠀⠀Formula for nth term :-

$\boxed{\bf{t_{n} = a_{1} + (n - 1)d}}$

Where :-

• $t_{n}$ = Nth term of the AP
• $a_{1}$ = First Term of the AP
• $n$ = No. of terms
• $d$ = Common Difference

Concept :-

Let the first 4 terms of the AP be a, (a + d) , (a + 2d) and (a + 3d).

⠀⠀⠀⠀So, According to the Question :-

$:\implies \bf{a + (a + d) + (a + 2d) + (a + 3d) = 142}$

So by solving this Equation , we will get the First Equation .

Now , we know that the 11th term of the AP is

(a + 10d) and the 3rd term is (a + 2d).

⠀⠀⠀So , According to the Question :-

$:\implies \bf{(a + 10d) = 2 + 2(a + 2d)}$

So , by solving this Equation , we will get the second Equation .

And then by solving this two Equations , we will get the required value.

Solution :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀Equation.(i)

Given Equation :-

$:\implies \bf{a + (a + d) + (a + 2d) + (a + 3d) = 142}$

By solving it , we get :-

$:\implies \bf{(a + a + a + a) + (d + 2d + 3d) = 142} \\ \\ \\ :\implies \bf{4a + 6d = 142} \\ \\ \\ \therefore \purple{\bf{4a + 6d = 142}}$

Hence, the Equation (i) is (4a + 6d = 142).

⠀⠀⠀⠀⠀⠀⠀⠀ Equation.(ii)

Given Equation :-

$:\implies \bf{(a + 10d) = 2 + 2(a + 2d)}$

By solving it , we get :-

$:\implies \bf{(a + 10d) = 2 + 2a + 4d} \\ \\ \\ :\implies \bf{(a + 10d) - (2a + 4d) = 2} \\ \\ \\ :\implies \bf{a + 10d - 2a - 4d = 2} \\ \\ \\ :\implies \bf{- a + 6d = 2} \\ \\ \\ \therefore \purple{\bf{- a + 6d = 2}}$

Hence , Equation (ii) is (- a + 6d = 2).

Now , putting the two Equations together , we get :-

⠀⠀⠀⠀⠀$\bf{4a + 6d = 142}$

⠀⠀⠀⠀⠀$\bf{- a + 6d = 2}$

⠀⠀⠀⠀⠀________________[By Subtracting]

⠀⠀⠀⠀⠀$\bf{5a = 140}$

$:\implies \bf{a = \dfrac{140}{5}} \\ \\ :\implies \bf{a = 28} \\ \\ \therefore \purple{\bf{a = 28}}$

Hence, the first term of the AP is 28.

Now , putting the value of first term (a) , in the equation (i) , we get :-

$:\implies \bf{4a + 6d = 142} \\ \\ :\implies \bf{4(28)+ 6d = 142} \\ \\ :\implies \bf{112 + 6d = 142} \\ \\ :\implies \bf{6d = 142 - 112} \\ \\ :\implies \bf{6d = 30} \\ \\ :\implies \bf{d = \dfrac{30}{6}} \\ \\ :\implies \bf{d = 5} \\ \\ \therefore \purple{\bf{d = 5}}$

Hence, the common difference of the AP is 5.

Answer 2

Step-by-step explanation:

d is 5 sure ... ok bye see y to tomorrow