Question

Solve the ques 2. ( iv )

Answer 1

HEY THERE!!


Let to be First term'a' and it's CommOn difference 'd' .

Let Sn it denotes Sum of terms of Number;-

According to the Question;-

Here, consisting of two Arithmetic progressions;-


S1=5+9+13+.....+77+81, (equation-1).
where First terms=5, common difference =9-5 =4

S2=-41-39-37-...(-3) (Equation 2)


See attachment;-


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Method Of Solution;-

Let to be First term'a' and it's CommOn difference 'd' .

Let Sn it denotes Sum of terms of Number;-

According to the Question;-

Sn= (3n²+6n) (Given)

Let Substitute n=n-1

S(n-1)=3(n-1)²+6(n-1)

=> 3(n²-2n+1)+6(n-1)

=> 3n²-6n+3+6n-6

=> 3n²-3

Hence, Value of Sn-1= 3n²-3n

We know that Formula of Summation of terms

Nth=Sn-Sn-1

Substitute the value of Required terms;-

nth=Sn-Sn-1

=> 3n²+6n -(3n²-3)

=> 3n²+6n-3n²+3

=> 6n+3

Hence, Nth term = 6n+3

Again, According to the question;-

Putting n=15 in Equation;-

6n+3

=>6(15)+3

=>90+3

=>93

Conclusion;-

Hence,the nth terms (6n+3) and the 15th terms is 93.

Answer 2

Hey there!!

To find :

5 + (-41) + 9 + (-39) + 13 + (-37) +......

Separate two Arthimetic progression differently.

= 5 + 9 + 13 +..... 81 + (-41) + (-39) +.. (-3)

To find sum S1 from 5, 9,13...81 :

S = n/2 (a + l)

Here a = 5, l = 81 , d = 9-5 = 4.

To find n:

l = a + (n-1)d

=) 81 = 5 + (n-1) 4

=) 76 = (n-1)4

=) 76/4 = (n-1)

=) 19 = (n-1)

=) 19 + 1 = n

=) 20 = n

S1 = 20/2 (5+81)

= 20/2 * 86

= 10* 86

= 860

To find sum S2 ; (-41) + (- 39) +.. (-3)

Here S2 = n/2 (a+l)

Here a = - 41, l = - 3
d = - 39 + 41 = 2

to find n:

=) l = a + (n-1)d

=) - 3 = - 41 + (n-1) 2

=) - 3 + 41 = (n-1) 2

=) 38 = (n-1)2

=) 38/2 = (n-1)

=) 19 = n-1

=) 19+1 = n

=) 20 = n

Hence S2 = 20/2* (-41 - 3)

= 10 (-44)

= - 440

Hence total = S1 + S2

= 860 + (-440)

= 420

Hope it helps uh!!