Q 21 :    

Let Tr be the rth term of an A.P. If for some mTm=125, T25=120 and 20r=125Tr=13, then 5mr=m2mTr is equal to          [2025]

  • 112

     

  • 126

     

  • 142

     

  • 98

     

(2)

We have, Tm=125, T25=120 and 20r=125Tr=13

Tm=a+(m1)d=125          ... (i)

and T25=a+24d=120          ... (ii)

Also, 20r=125Tr=13

 20[252(2a+24d)]=13

 20[252(a+120)]=13           (Using eqn. (ii))

 a=1500, d=1500

Using (i), we get m = 20

Now, 5mr=m2mTr=5×20r=2040Tr=126.



Q 22 :    

Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :          [2025]

  • 84

     

  • 90

     

  • 108

     

  • 122

     

(2)

We have, a1+a2+a3=54

 3(a+d)=54  a+d=18

        S20=10[2a+19d]=10[36+17d]

As   1600<S20<1800

 1600<10(36+17d)<1800

 160<36+17d<180

 124<17d<144

 7517<d<8817

Since, A.P. is of positive integers.

Common difference will be a natural number

  d=8  a=10

  a11=10+10×8=90.



Q 23 :    

The roots of the quadratic equation 3x2px+q=0 are 10th and 11th terms of an arithmetic progression with common difference 32. If the sum of the first 11 terms of this arithmetic progression is 88, then q – 2p is equal to __________.          [2025]



(474)

We have, S11=112(2a+10d)=88

 a+5d=8

 a=85×32=12          [ d=32]

Let α, β are the roots of the given quadratic equation.

  α=T10=a+9d=12+9×32=14

and β=T11=a+10d=12+10×32=312

Sum of roots = p3=T10+T11=14+312=592  p=1772

and product of roots = q3=T10×T11=7×31=217

 q=651.

Now, q – 2p = 651 – 177 = 474.



Q 24 :    

The interior angles of a polygon with n sides, are in an A.P. with common difference 6o. If the largest interior angle of the polygon is 219o, then n is equal to __________.          [2025]



(20)

Sum of interior angles =(n2)×180o

 n2(2a+(n1)6)=(n2)×180

 an+3n23n=(n2)×180          ... (i)

Now, according to question a + (n – 1)6 = 219

 a=2256n          ... (ii)

Putting value of a from equation (ii) to (i), we get

225n6n2+3n23n=180n360

 3n242n360=0

 n214n120=0

 n=20,6  (Rejected)



Q 25 :    

Let a1,a2,...,a2024 be an Arithmetic Progression such that a1+(a5+a10+a15+...+a2020)+a2024=2233. Then a1+a2+a3+...+a2024 is equal to __________.          [2025]



(11132)

We have,

a1+a5+a10+a15+...+a2020+a2024=2233          ... (i)

Since, a1,a2,...,a2024 be in A.P.

   a1+a2024=a5+a2020=a10+a2015.....

                         {Sum of terms equidistant form ends is equal}

From (i), we get

(a1+a2024)+(a5+a2020)+(a10+a2015)+...=2233

 203(a1+a2024)=2233          [ Total pairs = 203]

 a1+a2024=11

   Sum of n terms in A.P. is

        Sn=n2(a1+a2024)

               =1012×11          [ n = 2024]

   Sn=11132.