Question

If there are 6 arithmetic means between 5 and 33 then the common difference is

Answer 1

Answer:

The shortcut method for solving this question

d(common diff)=b-a/n+1=33-5/6+1=4

where a denotes the first term which is given

and b denotes the last term given

and n=no. of arithmatic mean5 years agoHelpfull: Yes(4) No(0)

No(0)There should be six terms between 5 and 33, hence there will be 7 differences

7 differences = 28

D= 4

The series is 5, 9, 13, 17, 21, 25, 29, 33

Answer 2

☯ $\normalsize\sf {\underline{\underline{Given:-}}}$

$\bullet \;\;\;\normalsize\sf{There\;are\;6\;arithmetic\;means\;b/w\;5\;and\;33}$

☯ $\normalsize\sf {\underline{\underline{To\;Find:-}}}$

$\bullet \;\;\;\normalsize\sf{Common\;difference\;of\;6\;arithmetic\;means\;b/w\;5\;and\;33}$

☯ $\normalsize\sf {\underline{\underline{Solution:-}}}$

$\bullet \;\;\;\normalsize\sf{Let's\;First\;term\;( a_1 ) = 5} \\\\ \bullet \;\;\;\normalsize\sf{Let's\;7th\;term\;( a_8 ) = 33}$

• $\normalsize\sf a_1 = 5$

• $\normalsize\sf a_2 = 5 + d$

• $\normalsize\sf a_3 = 5 + 2d$

• $\normalsize\sf a_4 = 5 + 3d$

• $\normalsize\sf a_5 = 5 + 4d$

• $\normalsize\sf a_6 = 5 + 5d$

• $\normalsize\sf a_7 = 5 + 6d$

• $\normalsize\sf a_8 = 33$

$\rule{200}{3}$

Now,

Lets $\normalsize\sf a_8 \; be\;n^{th}\;term\;of\;AP$

$\dashrightarrow\normalsize\sf a_8 = a + (n - 1)d \\\\ \dashrightarrow\normalsize\sf 33 = 5 + (8 - 1)d \\\\ \dashrightarrow\normalsize\sf 33 = 5 + 7d \\\\ \dashrightarrow\normalsize\sf 33 - 5 = + 7d \\\\ \dashrightarrow\normalsize\sf 28 = 7d \\\\ \dashrightarrow\normalsize\sf d = \cancel{ \dfrac{28}{7}} \\\\ \dashrightarrow\normalsize {\underline{\sf{\red{d = 4}}}}$

$\bullet \;\;\;\normalsize\sf Common\;difference\;of\;6\;arithmetic\;means\;b/w\;5\;and\;33\;is\;4$

$\;\;\;\;\;\;\;\;\;\;\; \normalsize{\underline{\underline{\sf{\dag\;Hence\;Solved!}}}}$

$\rule{200}{3}$