Question

guys plz solve this question, plz explain and write steps also of few !​

Answer 1

Answer:

a31=?

a11=38

a16=73

a31=a+30d

a31=-32+30(?)

=-32+210

=178

a+19d=38

a+15d=73

a=-32

d=7

Answer 2

i) We know if a, b, c are in AP, then

$\star\sf\;\; b = \dfrac{a + c}{2}$

Here,

⠀⠀⠀⠀⠀⠀⠀

AP = 2, $\sf \boxed{ \: ? \: }$ , 26

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf \boxed{\: ? \:} = \dfrac{2 + 26}{2} = \dfrac{28}{2} = \bf{14}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\sf ii) \:Here\;,\;AP = \boxed{\: ? \:}\;,\;13\; ,\; \boxed{\: ? \:}\;, 3$

$:\implies\sf \boxed{\: ? \:}\;,\; 13\;,\; \boxed{ \dfrac{13 + 3}{2}}\;,\;3$

$:\implies\sf \boxed{\: ? \:}\;,\;13\;,\; \boxed{\:8\:}\;,\; 3$

We know that,

⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow\sf d = a_3 - a_2$

$\dashrightarrow\sf d = 8 - 13 = - 5$

Also,

⠀⠀⠀⠀⠀⠀⠀

$\dashrightarrow\sf a_2 - a_1 = d$

$\dashrightarrow\sf 13 - a_1 = - 5$

$\dashrightarrow\sf a_1 = 13 + 5$

$\dashrightarrow\sf a_1 = 18$

Therefore,

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf \boxed{\: 18 \:}\;,\; 13\;, \;\boxed{\:8\:}\;,\; 3$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\sf iii) \:Here\;,\;AP = 5\;, \;\boxed{\: ? \:}\;,\; \boxed{\: ? \:}\;, \;9 \dfrac{1}{2}$

• First term, a = 5

• n = 4

• $\sf a_4 = 9 \dfrac{1}{2} = \dfrac{19}{2}$

• Common difference, d = ?

⠀⠀⠀⠀⠀⠀⠀

We know that,

⠀⠀⠀⠀⠀⠀⠀

$\sf a_4 = a + 3d$

$:\implies\sf \dfrac{19}{2} = 5 + 3d$

$:\implies\sf 3d = \dfrac{19}{2} - \dfrac{5}{1}$

$:\implies\sf 3d = \dfrac{19 - 10}{2} = \dfrac{9}{2}$

$:\implies\sf d = \dfrac{9}{2} \times \dfrac{1}{3} = \dfrac{3}{2}$

Therefore,

⠀⠀⠀⠀⠀⠀⠀

$\sf a_2 = a + d = 5 + \dfrac{3}{2} = \dfrac{10}{3} = \dfrac{13}{2} = 6 \dfrac{1}{2}$

$\sf a_3 = a + 2d = 5 + 2 \times \dfrac{3}{2} =5 + 3 = 8$

$\therefore\;\sf Hence,\;the \;required\; AP\; is,\; 5\;, \;\boxed{\: 6 \dfrac{1}{2} \:}\;,\; \boxed{\: 8 \:}\;, \;9 \dfrac{1}{2}$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\sf iv) \:Here\;,\;AP = -4\;, \;\boxed{\:?\:}\;,\; \boxed{\:?\:}\;,\;\boxed{\:?\:}\;,\; \boxed{\:?\:}\;,\; 6$

• First term, a = - 4

• n = 6

• d = ?

• $\sf a_6$ = 6

⠀⠀⠀⠀⠀⠀⠀

We know that,

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a_6 = a + 5d$

$:\implies\sf 6 = - 4 + 5d$

$:\implies\sf 5d = 6 + 4$

$:\implies\sf 5d = 10$

$:\implies\sf d = \cancel{ \dfrac{10}{5}}$

$:\implies\sf d = 2$

Therefore,

⠀⠀⠀⠀⠀⠀⠀

• $\sf a_2 = a + d = - 4 + 2 = - 2$

• $\sf a_3 = a + 2d = - 4 + 2 \times 2 = - 4 + 4 = 0$

• $\sf a_4 = a + 3d = - 4 + 3 \times 2 = - 4 + 6 = 2$

• $\sf a_5 = a + 4d = - 4 + 4 \times 2 = - 4 + 8 = 4$

⠀⠀⠀⠀⠀⠀⠀

$\therefore\;\sf Hence,\;the \;required\; AP\; is,\; -4\;,\; \boxed{\:-2\:}\;,\; \boxed{\:0\:}\;,\; \boxed{\:2\:}\;,\; \boxed{\:4\:}\;,\; 6$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\sf v) \:Here\;,\;AP = \boxed{\:?\:}\;,\; 38, \boxed{\:?\:}\;,\; \boxed{\:?\:}\;,\;\boxed{\:?\:}\;,\; - 22$

Given,

⠀⠀⠀⠀⠀⠀⠀

• $\sf a_2 = 38$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a + d = 38\qquad\qquad\bigg\lgroup\bf eq\;(1)\bigg\rgroup$

• $\sf a_6 = - 22$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf a + 5d = - 22\qquad\qquad\bigg\lgroup\bf eq\;(2)\bigg\rgroup$

Subtracting eq (1) from eq (2),

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 4d = - 60$

$:\implies\sf d = \cancel{-60}{4}$

$:\implies\sf d = - 15$

Now, Putting the value of d in eq (1),

$:\implies\sf a - 15 = 38$

$:\implies\sf a = 38 + 15 = 53$

Therefore,

⠀⠀⠀⠀⠀⠀⠀

• $\sf a_1 = 53$

• $\sf a_3 = a + 2d = 53 + 2(-15) = 53 - 30 = 23$

• $\sf a_4 = a + 3d = 53 + 3(-15) = 53 - 45 = 8$

• $\sf a_5 = a + 4d = 53 + 4(-15) = 53 - 60 = - 7$

⠀⠀⠀⠀⠀⠀⠀

$\therefore\;\sf Hence,\;the \;required\; AP\; is,\; \boxed{\:53\:}\;,\;38, \boxed{\:23\:}\;,\;\boxed{\:8\:}\;,\; \boxed{\:-7\:}\;,\; -22$

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

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

⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

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