Question

sum of natural numbers from 1 to N is 36 find value of n​

Answer 1

Pls mark brainliest :D

Have a gr8 day!!!

Answer 2

Answer:

Question :

The sum of natural numbers from 1 to n is 36. Find value of n.

$\begin{gathered}\end{gathered}$

Given :

• ➞ Sum of n terms = 36
• ➞ Common differences = 1
• ➞ First term = 1

$\begin{gathered}\end{gathered}$

To Find

• ➞ Value of n

$\begin{gathered}\end{gathered}$

Using Formula :

${\longrightarrow{\small{\underline{\boxed{\sf{\red{{S_n} = \dfrac{n}{2} \bigg[2a + \bigg(n - 1 \bigg)d\bigg]}}}}}}}$

• ➟ $\sf{S_n}$ = Sum of n terms
• ➟ $\sf{a}$ = First term
• ➟ $\sf{d}$ = common difference

$\begin{gathered}\end{gathered}$

Solution :

Finding the value of n by substituting the values in the formula :

${:\implies{\small{\sf{{S_n} = \dfrac{n}{2} \bigg[2a + \bigg(n - 1 \bigg)d\bigg]}}}}$

Substituting the given values

${:\implies{\small{\sf{36 = \dfrac{n}{2} \bigg[2 \times 1 + \bigg(n - 1 \bigg)1\bigg]}}}}$

${:\implies{\small{\sf{36 = \dfrac{n}{2} \bigg[2 + \bigg(n - 1 \bigg)1\bigg]}}}}$

Multiplying (n - 1) by 1

${:\implies{\small{\sf{36 = \dfrac{n}{2} \bigg[2 + \bigg(n \times 1 - 1 \times 1\bigg)\bigg]}}}}$

${:\implies{\small{\sf{36 = \dfrac{n}{2} \bigg[2 + \bigg(n - 1\bigg)\bigg]}}}}$

Now, taking RHS (2) to LHS and opening round brackets

${:\implies{\small{\sf{36 \times 2 = n \bigg[2 + n - 1\bigg]}}}}$

${:\implies{\small{\sf{72 = n \bigg[2 + n - 1\bigg]}}}}$

Now, subtracting 1 from 2

${:\implies{\small{\sf{72 = n \bigg[1 + n \bigg]}}}}$

Now, multiplying n into 1 + n

${:\implies{\small{\sf{72 = \bigg[1 \times n + n \times n \bigg]}}}}$

${:\implies{\small{\sf{72 = \bigg[ n + {n}^{2} \bigg]}}}}$

Opening square brackets

${:\implies{\small{\sf{72 = n + {n}^{2}}}}}$

Now, making a equation

${:\implies{\small{\sf{{n}^{2} + n - 72 = 0}}}}$

${:\implies{\small{\sf{{n}^{2} + 9n - 8n - 72 = 0}}}}$

${:\implies{\small{\sf{ n(n + 9) - 8(n + 9) = 0}}}}$

${:\implies{\small{\sf{(n - 8)(n + 9) = 0}}}}$

Solving for (n - 8) = 0

${:\implies{\small{\sf{n - 8 = 0}}}}$

${:\implies{\small{\sf{n = 0 + 8}}}}$

${\therefore{\sf{\red{ \: n = 8}}}}$

Solving for (n + 9) = 0

${:\implies{\small{\sf{n + 9 = 0}}}}$

${:\implies{\small{\sf{n = 0 - 9}}}}$

${\therefore{\sf{\red{n = - 9}}}}$

Since, the number of terms n can’t be negative.

Hence, The value of n is 8.

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Learn More :

★ Formula to find the numbers of term of an AP:

$\longrightarrow{\small{\underline{\boxed{\sf{\purple{n= \bigg[ \dfrac{(l - a)}{d} \bigg]}}}}}}$

★ Formula to find the tsum of first n terms of an AP:

$\longrightarrow{\small{\underline{\boxed{\sf{\purple{S_n= \dfrac{n}{2} \big(a + l \big)}}}}}}$

★ Formula to find the sum of squares of first n natural numbers of an AP:

$\longrightarrow{\small{\underline{\boxed{\sf{\purple{S = \dfrac{n(n + 1)(2n + 1)}{6} }}}}}}$

★ Formula to find the nth term of an AP is the square of the number of terms:

$\longrightarrow{\small{\underline{\boxed{\sf{\purple{S = {(n)}^{2} }}}}}}$

★ Formula to find the sum of of an AP:

$\longrightarrow{\small{\underline{\boxed{\sf{\purple{S = n(n+1)}}}}}}$

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