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Title : Qa As Well As Di | Query No.-57 | Xat 2016 Paper
link : Qa As Well As Di | Query No.-57 | Xat 2016 Paper

Qa As Well As Di | Query No.-57 | Xat 2016 Paper

Actual Question of XAT

Question is: f is a purpose for which f(1)=1 in addition to f(x)=2x+f(x-1) for each natural pose out x≥2. Find f(31).

Options are:

Option(a): 869

Option(b): 929

Option(c): 951

Option(d): 991

Option(e): None of the above.

Please Solve Sincerely!

Solution is:

We convey given, f(1) = 1 in addition to f(x)=2x+f(x-1) where x≥2 , f(31)=?

if nosotros pose x=1 in f(x)=2x+f(x-1) then

f(2) =2*2+f(2-1) = 4+f(1) = 4+1 = 5

So, instantly nosotros convey f(1)=1 in addition to f(2)=5

f(3) =2*3+f(3-1) = 6+f(2) = 6+5 = 11

f(4) =2*4+f(4-1) = 8+f(3) = 8+11 = 19

f(5) =2*5+f(5-1) = 10+f(4) = 10+19 = 29

f(6) =2*6+f(6-1) = 12+f(5) = 12+29 = 41

f(7) =2*7+f(7-1) = 14+f(6) = 14+41 = 55

f(8) =2*8+f(8-1) = 16+f(7) = 16+55 = 71

f(9) =2*9+f(9-1) = 18+f(8) = 18+71 = 89

f(10) =2*10+f(10-1) = 20+f(9) = 20+89 = 109

f(11) =2*11+f(11-1) = 22+f(10) = 22+109 = 131

f(12) =2*12+f(12-1) = 24+f(11) = 24+131 = 155

f(13) =2*13+f(13-1) = 26+f(12) = 26+155 = 181

f(14) =2*14+f(14-1) = 28+f(13) = 28+181 = 209

f(15) =2*15+f(15-1) = 30+f(14) = 30+209 = 239

f(16) =2*16+f(16-1) = 32+f(15) = 32+239 = 271

f(17) =2*17+f(17-1) = 34+f(16) = 34+271 = 305

f(18) =2*18+f(18-1) = 36+f(17) = 36+305 = 341

f(19) =2*19+f(19-1) = 38+f(18) = 38+341 = 379

f(20) =2*20+f(20-1) = 40+f(19) = 40+379 = 419

f(21) =2*21+f(21-1) = 42+f(20) = 42+419 = 461

f(22) =2*22+f(22-1) = 44+f(21) = 44+461 = 505

f(23) =2*23+f(23-1) = 46+f(22) = 46+505 = 551

f(24) =2*24+f(24-1) = 48+f(23) = 48+551 = 599

f(25) =2*25+f(25-1) = 50+f(24) = 50+599 = 649

f(26) =2*26+f(26-1) = 52+f(25) = 52+649 = 701

f(27) =2*27+f(27-1) = 54+f(26) = 54+701 = 755

f(28) =2*28+f(28-1) = 56+f(27) = 56+755 = 811

f(29) =2*29+f(29-1) = 58+f(28) = 58+811 = 869

f(30) =2*30+f(30-1) = 60+f(29) = 60+869 = 929

f(31) =31*2+f(31-1) = 62+f(30) = 62+929= 991

Hence, f(31) =991 , Option(d) is correct.

Another Solution:

We convey f(1)=1, f(2) =5

f(31) =62+60+----------------------------+6 + 5

Now, uncovering the value of 62+60+----------------------------+6 yesteryear utilise addition of AP formula.

Sum of AP = n/2[2a+(n-1)d]

Find the pose out of damage inwards 62+60+----------------------------+6

We tin hand the sack write this AP Series as: 6+8+........................+60+62

nth term of AP, tn = a+(n-1)d 

where tn= concluding term = 62

a=first term = 6

d= mutual departure = 2

n= pose out of damage = ?

62 =6+(n-1)2

62 - vi = 2n-2

2n=62-6+2 = 58

n=58/2=29 = pose out of damage inwards the series.

Sum of AP = n/2[a+tn]

=29/2 * [6+62]

=29/2 * 68

=29*34

=986

f(31) =62+60+----------------------------+6 + five = 986+5 = 991

Hence, f(31) = 991, Option(d) is correct.

Happy Learning | BhakLol.Com

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