Question

plz answer Q no. 1,2,3and4

Answer 1

Hey Vansh ,

Giving solutions of all questions at one time is difficult.

I am giving the answer of all and some hint .

Q.1)replace n by (2m-1) we get ratio of MTH term i.e
tm/tm*=[5(2m-1)-3]/[4(2m-1)+18]
=(10m-8)/(8m+14) .

Now replace m by 11 we get,
T11/T11*=102/102=1.

Ans :=>1:1.

Q.3l et PR =x+3,PT=x+2, ,RT =x.

Perimeter=26.

Then ,x=7.(find),PR=10,PT=9,RT=7.

Also,let PQ=PT=a,QR=RS=b,ST=TU=c

Then ,adding all three and finding we will get =>a=6,b=4,c=3.

So,QR+RT=b+b+c=8+3=11.

Ans :=>11.

Q.4) AM =sum of all terms of AP/no of terms =36/4=9.

Ans :=>9.

Q.2 ) Nice question


[(6-x)/x]^9=[(5+1-x)/x]^9

={[(5^1/3)^3+1^3-x]/x}^9

=>{(5^1/3 +1)[5^(2/3)+1-(5^1/3) ]-x/x }^9
(Using a^3+b^3=(a+b)(a^2+b^2-ab) )

=>{[[{5^(1/3)+1}x ]-x]/x}^9

=(5^1/3)^9=(5)^1/3×9=5^3=125

Ans:=>125.

Hope it will help you a lot.
#kkashyap