Question

In an arithmetic sequence

S10 = 125, a1 = 17 find the common difference, d,of the sequence​

Answer 1

$\bf\huge\blue{\underline{\underline{ Question : }}}$

In an arithmetic sequence S10 = 125, a1 = 17 find the common difference(d) & the sequence.

$\bf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• S10 = 125
• a = 17
• n = 10

★ To find,

• Common difference (d).
• AP series.

★ Formula used:

$\tt\red{:\implies S_{n} = \frac{n}{2} [ 2a + (n - 1)d ]}$

• Substitute the values.

$\bf\:\leadsto 125 = \frac{10}{2} [ 2(17) + (10 - 1)d ]$

$\bf\:\leadsto 125 = 5 [ 34 + (9)d ]$

$\bf\:\leadsto \frac{125}{5} = 34 + 9d$

$\bf\:\leadsto 25 = 34 + 9d$

$\bf\:\leadsto 25 - 34 = 9d$

$\bf\:\leadsto 9d = - 9$

$\bf\:\leadsto d = \frac{- 9}{9}$

$\bf\:\leadsto d = - 1$

$\underline{\boxed{\rm{\purple{\therefore Hence,\:common\:difference\:is-1}}}}\:\orange{\bigstar}$

★ Now,

We know that, the general form of an AP series is :-

➡ a, a + d, a + 2d, a + 3d....

• Substitute the values.

$\bf\:\leadsto 17, 17 + (-1),17 + 2(-1), 17 + 3(-1)....$

$\bf\:\leadsto 17, 17 - 1,17 - 2, 17 - 3....$

$\bf\:\leadsto 17, 16,15,14....$

$\underline{\boxed{\rm{\purple{\therefore Hence,\:the\:AP\:series \:is\: 17,16,15,14,....}}}}\:\orange{\bigstar}$

★ Verification,

• Substitute all the values in the formula.

$\bf\:\leadsto S_{10} =\frac {10}{2} [ 2(17) + (10 - 1)(-1)]$

$\bf\:\leadsto S_{10} =5 [ 34 + (9)(-1)]$

$\bf\:\leadsto S_{10} =5 [ 34 - 9]$

$\bf\:\leadsto S_{10} =5 [25]$

$\bf\:\leadsto S_{10} =125$

#Hence,it was verified.

★ More information,

$\boxed{\begin{minipage}{5 cm} AP Formulae \\ \\ : \implies a_{n} = a + (n - 1)d \\ \\ :\implies S_{n} = \frac{n}{2} [ 2a + (n - 1)d ] \end{minipage}}$

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