Question

answer for 6question

Answer 1

Answer:

Hey there !

Solution:

\begin{gathered} Sum \ of \ n \ terms \ of \ an \ AP \ can \ be \ given \ as: \\\\ S_n = \frac{n}{2} ( a + a_n ). \\\\ Here \ a \ represents \ the \ first \ term \ and \ a_n \ represents \ the \ last \ term. \\\ Substituting \ the \ given \ values \ we \ get, \\\\ 57 = \frac{n}{2} ( 17 + 2 ) \\\\ = > 57 ( 2 ) = n ( 17 + 2 ) \\\\ = > 114 = 19 \ n \\\\ = > n = \frac{114}{19} = 6. \\\\ \boxed{ Hence \ number \ of \ terms \ ( n ) \ is \ 6} \end{gathered}

Sum of n terms of an AP can be given as:

S

n

=

2

n

(a+a

n

).

Here a represents the first term and a

n

represents the last term.

Substituting the given values we get,

57=

2

n

(17+2)

=>57(2)=n(17+2)

=>114=19 n

=>n=

19

114

=6.

Hence number of terms (n) is 6

\begin{gathered} Now, \ we \ know \ a_n, \ a, \ n. \\\\ Using \ this \ we \ can \ substitute \ in \ another \ formula \ to \ get \ common \ difference. \\\\ Formula \ to \ be \ applied: \\\\ a_n = a + ( n - 1 ) d \\\\ Substituting\ the\ values\ for\ everything\ we\ get \\\\ = > 17 = 2 + ( 6 - 1 ) d \\\\ = > 17 - 2 = (5)d \\\\ = > 15 = 5d \\\\ = > d = \frac{15}{5} = 3 \\\\ { \boxed{Hence \ the \ common \ difference \ ( d ) \ is \ 3}} \end{gathered}

Now, we know a

n

, a, n.

Using this we can substitute in another formula to get common difference.

Formula to be applied:

a

n

=a+(n−1)d

Substituting the values for everything we get

=>17=2+(6−1)d

=>17−2=(5)d

=>15=5d

=>d=

5

15

=3

Hence the common difference (d) is 3

Hope my answer helped !