Question

very important question solve.
Solve the equation:
1 + 4 + 7 + 10 + ... + x = 287​

Answer 1

$\large\underline{\sf{Solution-}}$

Given series is

$\rm \: 1 + 4 + 7 + 10 + ... + x = 287$

Since,

$\rm \: \: \: \: \: 4 - 1 = 3$

$\rm \: \: \: \: \: 7 - 4 = 3$

$\rm \: \: \: \: \: 10 - 7 = 3$

$\rm\implies \:1 + 4 + 7 + 10 + ... \: forms \: an \: AP \: series$

whom

• First term, a = 1

• Common difference, d = 3

Now, given that

$\rm \: 1 + 4 + 7 + 10 + ... + x = 287$

Let assume that number of terms in the series be n.

It means, we have

$\rm \: \: \: \: \: \: a_1 = 1$

$\rm \: \: \: \: \: \: a_n = x$

$\rm \: \: \: \: \: \: S_n = 287$

$\rm \: \: \: \: \: \: d = 3$

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of AP.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

So, on substituting the values, we get

$\rm \: \dfrac{n}{2} \bigg(2 \times 1 + (n - 1) \times 3\bigg) = 287$

$\rm \: \dfrac{n}{2} \bigg(2+ 3n - 3\bigg) = 287$

$\rm \: \dfrac{n}{2} \bigg(3n - 1\bigg) = 287$

$\rm \: {3n}^{2} - n = 574$

$\rm \: {3n}^{2} - n - 574 = 0$

$\rm \: {3n}^{2} + 41n - 42n - 574 = 0$

$\rm \: n(3n + 41) - 14(3n + 41) = 0$

$\rm \: (3n + 41)(n - 14) = 0$

$\rm\implies \:n \: = \: 14 \: \: or \: \: n = \: - \: \frac{41}{3} \: \{rejected \}$

Now,

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the progression.

• n is the no. of terms.

• d is the common difference.

As, given that

$\rm \: a_n = x$

means,

$\rm \: a_{14} = x$

$\rm \: a + 13d = x$

$\rm \: 1 + 13 \times 3 = x$

$\rm \: 1 + 39 = x$

$\bf\implies \:x = 40$

Answer 2

Answer:

Here,given = 1

d = 4 −1 = 3

And , sn =287

Now,

sn=2n(2a+(n−1)d)

⇒287=2n(2×1+(n−1)3)

⇒287=2n(2+3n−3)

⇒574=n(3n−1)

⇒574=3n2−n⇒3n2−n−574=0

on solving the quadratic equaton using formula n=2a−b±b2−4ac

We get n=14 & 3−41[does not exist] so, n=14

Now,

sn=2n(a+1)

⇒287=214(1+x)

⇒574=14(1+x)

⇒(1+x)=14574⇒1+x=41

⇒x=41−1

∴x=40

x=40 is the solution.