Question

$\mathfrak{\huge{\underline{Question:}}}$

The first and last term of an AP are 17 and 350 respectively. if the common difference is 9 how many terms are there and what is their terms?​

Answer 1

• let a and d be the first term and common difference for an AP.

• number of terms of AP = n

last term = nth term = l

given:

• a = 17
• d = 9 ,
• l = 350

a + ( n - 1 ) d=nth term

=> 35017 + ( n - 1 ) 9 = 350

=>( n - 1 ) 9 = 350 - 17

=>( n - 1 ) 9 = 333

=>n - 1 = 333 /9

=>n - 1 = 37

=>n = 37 + 1

=>n = 38

Therefore ,

number of terms in given AP = $\red{\bold{n=38}}$

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The terms are:-

• $\red{17, 26, 35....350}$

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Additional:

How to find the sum of the terms of AP?

let Sum of n terms of AP = Sn

we know,

Sn = n /2 ( a + l )

• here n= 38 (as we calculated)
• putting values we get:

Sum = 38 / 2 [ 17 + 350 ]

= 19 × 367

= 6973

Answer 2

AnswEr :

$\bf{Given}\begin{cases}\sf{First \ term \ (a) = 17}\\\sf{Common \ difference \ (d) = 9}\\ \sf{Last \ term \ (a_{n})= 350}\end{cases}$

${\dag}\underline{\frak{Using \ Arithmetic \ progression \ formula \ :}}$

$\star\:\small\boxed{\sf{\purple{a_{n} = a + [n - 1] d}}}$

$:\implies\sf 17 + (n - 1)9 = 350 \\\\\\:\implies\sf (n - 1)9 = 350 - 17 \\\\\\:\implies\sf (n - 1)9 = 333 \\\\\\:\implies\sf n - 1 = \dfrac{\cancel{333}}{\cancel{9}} \\\\\\:\implies\sf n - 1 = 37 \\\\\\:\implies\sf n = 37 + 1 \\\\\\:\implies\boxed{\sf{\purple{ n = 38}}}}$

$\therefore\underline{\sf{Here, \ we \ get \ n \ is \ 38.}}$

There are 38 terms in AP.

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${\dag}\underline{\frak{To \ find \ sum \ of \ an \ Arithmetic \ progression \ formula \ is \ given \ as \ : }}$

$\star\:\boxed{\sf{\purple{\Bigg(S_{n} = \dfrac{n}{2} (a + l) \Bigg)}}}$

$:\implies\sf S_{38} = \dfrac{\cancel{38}}{\cancel{2}} = (17 + 350) \qquad \quad \bigg\lgroup\bf n = 38 \bigg\rgroup\\\\\\:\implies\sf 19 \times 367 \\\\\\:\implies\boxed{\sf{\purple{S_{n} = 6973}}}$

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⠀⠀⠀$\boxed{\bf{\mid{\overline{\underline{\bigstar\: Used \ Formulas \ : }}}}\mid}$

$\begin{lgathered}\boxed{\begin{minipage}{15 em} \sf \displaystyle \bullet a_n=a + (n-1)d \\\\\\ \bullet S_n= \dfrac{n}{2} \left(a + a_n\right) \end{minipage}}\end{lgathered}$