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151.	The value of `0. 2^(logsqrt(5)1/4+1/8+1/(16)+)`is`4`b. `log4`c. `log2`d. none of theseA. 4B. log 4C. log 2D. none of these
Answer» Correct Answer - A We have, `1/4+1/8+1/16+…=(1//4)/(1-1//2)=1/2` Hence, `0.2^(log_(sqrt5(1/4+1/8+1/16+..)))=0.2^(log_(sqrt5)""1/2)` `=(1/5)^(log_(sqrt5)""1/2)` `=(5^(-1))^(2log_(5)""1/2)` `(5)^(-2log_(5)""1/2)` `=(5)^(log_(5)4)=4`

152.	If `(1+3+5++p)+(1+3+5++q)=(1+3+5++r)`where each set of parentheses contains the sum of consecutive oddintegers as shown, the smallest possible value of `p+q+r(w h e r ep >6)`is`12`b. `21`c. `45`d. `54`A. 12B. 21C. 45D. 54
Answer» Correct Answer - B We know that 1+3+5+….+(2k-1)=`k^(2)`. Thus, the given equation can be written as `((p+1)/2)^(2)+((q+1)/2)^(2)=((r+1)/2)^(2)` `rArr(p+1)^(2)+(q+1)^(2)=(r+1)^(2)` As `pgt6,p+1gt7`, we may take p+1=8,q+1=6, `r+1=10`. Hence, p+q+r=21

153.	If `t_n` denotes the nth term of the series 2+3+6+11+18+….. Then `t_50` isA. `49^2-1`B. `49^2`C. `50^2+1`D. `49^2+2`
Answer» Correct Answer - D `2+3+6+11+18+…` `=(0^(2)+2)+(1^(2)+2)+(2^(2)+2)+(3^(2)+2)+…` Hence, `t_(50)=49^(2)+2`.

154.	Two arithmetic progressions have the same numbers. The reatio of the last term of the first progression to the first term of the second progression is equal to the ratio of the last term of the second progression to the first term of first progression is equal to 4. The ratio of the sum of the n terms of the first progression to the sum of the n terms of teh first progression to the sum of the n terms of the second progerssion is equal to 2.A. `6//5`B. `7//2`C. `9//5`D. none of these
Answer» Correct Answer - B Let the first term be a and common difference be d for the first A.P. and the first term be b and common difference be e for the second A.P. and let the numbers of terms be n. Then `(a+(n-1)d)/b=(b+(n-1)e)/a=4` (1) `(n/2[2a+(n-1)d])/(n/2[2b+(n-1)e])=2` (2) From (1) and (2), we get a-4b+(n-1)d=0 (3) b-4a+(n-1)e=0 (4) 2a-4b+(n-1)d-2(n-1)e=0 (5) `4xx(3)+(4)` gives `-15b+4(n-1)d+(n-1)e=0` (6) `(4)+2xx(5)` gives `-7b+2(n-1)d-3(n-1)e=0` (7) Further, `15xx(7)-7xx(6)` gives `2(n-1)d-52(n-1)e=0` or d=26e `(becausengt1)` `therefored//e=26` Putting d=26e in (3) and solving it with (4), we get a=2(n-1)e,b=7(n-1)e Then, the ratio of their `n^(th)` terms is `(2(n-1)+(n-1)26e)/(7(n-1)e+(n-1)e)=7/2`

155.	Two arithmetic progressions have the same numbers. The reatio of the last term of the first progression to the first term of the second progression is equal to the ratio of the last term of the second progression to the first term of first progression is equal to 4. The ratio of the sum of the n terms of the first progression to the sum of the n terms of teh first progression to the sum of the n terms of the second progerssion is equal to 2. The ratio of their first term isA. `2//7`B. `3//5`C. `4//7`D. `2//5`
Answer» Correct Answer - A Let the first term be a and common difference be d for the first A.P. and the first term be b and common difference be e for the second A.P. and let the numbers of terms be n. Then `(a+(n-1)d)/b=(b+(n-1)e)/a=4` (1) `(n/2[2a+(n-1)d])/(n/2[2b+(n-1)e])=2` (2) From (1) and (2), we get a-4b+(n-1)d=0 (3) b-4a+(n-1)e=0 (4) 2a-4b+(n-1)d-2(n-1)e=0 (5) `4xx(3)+(4)` gives `-15b+4(n-1)d+(n-1)e=0` (6) `(4)+2xx(5)` gives `-7b+2(n-1)d-3(n-1)e=0` (7) Further, `15xx(7)-7xx(6)` gives `2(n-1)d-52(n-1)e=0` or d=26e `(becausengt1)` `therefored//e=26` Putting d=26e in (3) and solving it with (4), we get a=2(n-1)e,b=7(n-1)e Then, the ratio of their `n^(th)` terms is `(2(n-1)+(n-1)26e)/(7(n-1)e+(n-1)e)=7/2`

156.	Two arithmetic progressions have the same numbers. The reatio of the last term of the first progression to the first term of the second progression is equal to the ratio of the last term of the second progression to the first term of first progression is equal to 4. The ratio of the sum of the n terms of the first progression to the sum of the n terms of teh first progression to the sum of the n terms of the second progerssion is equal to 2. The ratio of their common difference isA. 12B. 24C. 26D. 9
Answer» Correct Answer - C Let the first term be a and common difference be d for the first A.P. and the first term be b and common difference be e for the second A.P. and let the numbers of terms be n. Then `(a+(n-1)d)/b=(b+(n-1)e)/a=4` (1) `(n/2[2a+(n-1)d])/(n/2[2b+(n-1)e])=2` (2) From (1) and (2), we get a-4b+(n-1)d=0 (3) b-4a+(n-1)e=0 (4) 2a-4b+(n-1)d-2(n-1)e=0 (5) `4xx(3)+(4)` gives `-15b+4(n-1)d+(n-1)e=0` (6) `(4)+2xx(5)` gives `-7b+2(n-1)d-3(n-1)e=0` (7) Further, `15xx(7)-7xx(6)` gives `2(n-1)d-52(n-1)e=0` or d=26e `(becausengt1)` `therefored//e=26` Putting d=26e in (3) and solving it with (4), we get a=2(n-1)e,b=7(n-1)e Then, the ratio of their `n^(th)` terms is `(2(n-1)+(n-1)26e)/(7(n-1)e+(n-1)e)=7/2`

157.	Two consecutive numbers from 1,2,3 …., n are removed .The arithmetic mean of the remaining numbers is 105/4 The sum of all numbersA. exceeds 1600B. is less than 1500C. lies between 1300 and 1500D. none of these
Answer» Correct Answer - B Let m and (m+1) be the removed numbers from 1,2,…,n. Then, the sum of the remaining numbers is n(n+1)/2-(2m+1). From given condition, `105/4=((n(n+1))/2-(2m+1))/((n-2))` or `2n^(2)-103n-8m+206`=0 Since n and m are integers, so n must be even. Let n=2k. Then, `m=(4k^(2)+103(1-k))/4` Since m is an integer, then 1-k must be divisble by 4. Let k=1+4t. Then we get n=8t+2 and `m=16t^(2)-95t+1`. Now, `1lemltn` `rArr1le16t^(2)-95t+1lt8t+2` Solving, we get t=6. Hence, n=50 and m=7 Hence, the removed numbers are 7 and 8. Also, sm of all numbers is 50(50+1)/2=1275.

158.	The greatest interger by which `1+Sigma_(r=1)^(30) rxxr !` is divisible isA. composite numberB. odd numberC. divisible by 3D. none of these
Answer» Correct Answer - D `rxxr!=(r+1-1)xxr!` `=(r+1)!-r!` `=V(r )-V(r-1)` `rArrsum_(r=1)^(30)r(r!)=V(31)-V(0)=(31)!-1` `rArr1+sum_(r=1)^(30)r(r!)=31!` which is divisble by 31, which are prime numbers.

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