Question

23rd term of an arithmetic sequence is 32 . 35 th term is 104. Which is the middle of first 35 term

Answer 1

GivEn:-

• 23rd term of an arithmetic sequence is 32.

• 35th term of an arithmetic sequence is 104.

To find:-

• Which is the middle of first 35 term.

SoluTion:-

★ $\bf{\underline{\red{\underline{According\;to\;question:-}}}}$

• $\sf a_{23}$ = 32

• $\sf a_{35}$ = 104

Therefore,

a + 22d = 32 --(1)

a + 35d = 104 --(2)

★ Now, Subtracting eq(1) from eq(2) -

We get,

$:\implies$ 12d = 72

$:\implies$ d = $\sf \cancel{ \dfrac{72}{12}}$

$:\implies\bf {\underline{\blue{d = 6}}}$

★ Now, Substituting value of d in eq(1) :-

$:\implies$ a + 22 × 6 = 32

$:\implies$ a + 132 = 32

$:\implies\bf {\underline{\blue{a = -100}}}$

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$\dag\;\sf{\underline{In\;First\;35\;term,\;18th\;term\;will\;be\;the\;middle\;most -}}$

$:\implies$ $\sf a_{18}$ = a + (n - 1)d

$:\implies$ $\sf a_{18}$ = -100 + (18 - 1)6

$:\implies$ $\sf a_{18}$ = -100 + 17 × 6

$:\implies$ $\sf a_{18}$ = -100 + 17 × 6

$:\implies$ $\sf a_{18}$ = -100 + 102

$:\implies$ $\sf a_{18}$ = 2

$\therefore\;{\underline{\sf{\red{2\;is\;the\;middle\;most\;term.}}}}$

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