Question

THE 17TH TERM OF AN A.P.EXCEEDS ITS 10TH TERM BY 7.FIND THE COMMON DIFFERENCE?

Answer 1

☺ Hello mate__ ❤

◾◾here is your answer...

Given, 17th term exceeds its 10th term by 7.

It means a17 = a10 + 7         ........eq(1)

Here, a17 is the 17th term and a10 is the 10th term of an AP.

Using formula an=a+(n−1)d,   to find nth term of arithmetic progression, we get

a17=a+(16)d      ..........eq (2)

a10=a+(9)d    .........eq(3)

Putting (2) and (3) in equation (1), we get

a+16d=a+9d+7

⇒7d=7

⇒d=7/7=1

I hope, this will help you.

Thank you______❤

✿┅═══❁✿ Be Brainly✿❁═══┅✿

Answer 2

${\underline{\underline{\frak{Given\::}}}}$

17TH Term is AP is 7 More Than 10TH Term...

${\underline{\underline{\frak{To\:Find\::}}}}$

Common difference?

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${\underline{\underline{\frak{Solution\::}}}}$

$\underline{\bigstar\:\boldsymbol{According\:to\:the\: Question\::}}$

$:\implies\sf a_{17} = a_{10} + 7$

$:\implies\sf a + 16d = a + 9d + 7$

$:\implies\sf 16d = 9d + 7$

$:\implies\sf 16d - 9d = 7$

$:\implies\sf 7d = 7$

$:\implies\sf d = \cancel{ \dfrac{7}{7}}$

$:\implies{\boxed{\sf{\purple{d = 1}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;Common\; difference\;of\;an\;AP\:is\; \bf{1}.}}}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\qquad\qquad\boxed{\underline{\underline{\bigstar \: \bf\:Formula\:Related\:to\:AP\:\bigstar}}}$

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$\sf (i)\;The\; n^{th}\;term\;of\;an\;AP\; = \; \red{a_n + (n - 1)d}$

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$\sf (ii)\;Sum\;of\;n\;term\;of\;an\;AP\; = \; \purple{S_n = \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup}$

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$\sf (iii)\;Sum\;of\;all\;terms\;of\;AP\;having\;last\:term\;as\;'l'\; = \; \pink{ \dfrac{n}{2}(a + l)}$[/tex]

Hope iT helps