Question

the first and the last term of an ap are 5 and 45 respectively if the sum of all its term is 400 find the common difference and the number of terms​

Answer 1

Step-by-step explanation:

First term, a=5

Last term, l=45

Common difference = d

Sum of AP of n terms = (n/2)(a+l)

Sum = 400

On substituting values of a & l,

We get, n=16

l=a+(n-1)d

On substituting values of a,l & n

We get, d= 8/3

Answer 2

$\sf{\large{\underline{\underline{EXPLANATION:}}}}$

★Given,,

$\sf{Given}\begin{cases}\sf{a=5}\\ \sf{a_{n}=45}\\ \sf{s_{n}=400}\end{cases}$

★To find,,

• Common difference,d=??

• Number of terms,n=??

★Therefore,, Using formula,,

$\mathtt{\boxed{\blue{</strong><strong>s_</strong><strong>{</strong><strong>n</strong><strong>}</strong><strong>=</strong><strong>\</strong><strong>f</strong><strong>r</strong><strong>a</strong><strong>c</strong><strong>{</strong><strong>n</strong><strong>}</strong><strong>{</strong><strong>2</strong><strong>}</strong><strong>[</strong><strong>a</strong><strong>+</strong><strong>a_</strong><strong>{</strong><strong>n</strong><strong>}</strong><strong>]</strong><strong>}}}$

$\</strong><strong>s</strong><strong>f</strong><strong>{\</strong><strong>i</strong><strong>m</strong><strong>p</strong><strong>l</strong><strong>i</strong><strong>e</strong><strong>s</strong><strong>{s_{n}=\frac{n}{2}[</strong><strong>5</strong><strong>+</strong><strong>4</strong><strong>5</strong><strong>]}}$

$\sf{\implies{</strong><strong>4</strong><strong>0</strong><strong>0</strong><strong>=\frac{n}{2}</strong><strong>\</strong><strong>t</strong><strong>i</strong><strong>m</strong><strong>e</strong><strong>s</strong><strong>{</strong><strong>5</strong><strong>0</strong><strong>}</strong><strong>}}$

$\sf{\implies{</strong><strong>\</strong><strong>c</strong><strong>a</strong><strong>n</strong><strong>c</strong><strong>e</strong><strong>l</strong><strong>\</strong><strong>d</strong><strong>f</strong><strong>r</strong><strong>a</strong><strong>c</strong><strong>{</strong><strong>4</strong><strong>0</strong><strong>0</strong><strong>}</strong><strong>{</strong><strong>2</strong><strong>5</strong><strong>}</strong><strong>=</strong><strong>n</strong><strong>}}$

$\sf{\implies{</strong><strong>1</strong><strong>6</strong><strong>=n}}$

• Therefore number of terms=16

★Now,,

$\mathtt{\boxed{\blue{</strong><strong>a_</strong><strong>{</strong><strong>n</strong><strong>}</strong><strong>=</strong><strong>a</strong><strong>+</strong><strong>(</strong><strong>n-1</strong><strong>)</strong><strong>d</strong><strong>}}}$

$\</strong><strong>s</strong><strong>f</strong><strong>{\</strong><strong>i</strong><strong>m</strong><strong>p</strong><strong>l</strong><strong>i</strong><strong>e</strong><strong>s</strong><strong>{</strong><strong>4</strong><strong>5</strong><strong>=</strong><strong>5</strong><strong>+(</strong><strong>1</strong><strong>6</strong><strong>-1)d}}$

$\sf{\implies{</strong><strong>4</strong><strong>5</strong><strong>-</strong><strong>5</strong><strong>=</strong><strong>1</strong><strong>5</strong><strong>d}}$

$\sf{\implies{</strong><strong>\</strong><strong>c</strong><strong>a</strong><strong>n</strong><strong>c</strong><strong>e</strong><strong>l</strong><strong>\</strong><strong>d</strong><strong>f</strong><strong>r</strong><strong>a</strong><strong>c</strong><strong>{</strong><strong>4</strong><strong>0</strong><strong>}</strong><strong>{</strong><strong>1</strong><strong>5</strong><strong>}</strong><strong>=</strong><strong>d</strong><strong>}}$

$\sf{\implies{\frac{</strong><strong>8</strong><strong>}{</strong><strong>3</strong><strong>}=d}}$

• Therefore common difference,d=8/3

$\rule{200}2$

Hope it helps ❣️❣️