Number Series Reasoning Test

Go on adding 7, 9, 11, 13, 15, 17 respectively to obtain the next number.

Go on dividing by 4 to get the next number.

Let no's be a,b,c,d,e,f,g. then if a=7;  b=a1+1; c=b2+2 and so on

Analyzing the differences between consecutive terms: 190-166=24, 166-145=21, 145-128=17, 128-112=16, 112-100=12, 100-91=9. The differences decrease by 3 (24,21,18,15,12,9), but 145-128=17 instead of 18. This breaks the pattern. Alternatively, looking at sums of digits: 190=10, 166=13, 145=10, 128=11, 112=4, 100=1, 91=10 - 128 is the only one not reducing to 1, 4, or 10.

Go on multiplying with 1, 2, 3, 4, 5, 6 to get the next number.

Go on subtracting 3 and divide the result by 2 to obtain the next number.

Multiply by 2 and add 1to get the next number.

Go on dividing by 6,5,4,3,2,1 respectively to obtain the next number.

Go on adding 7, 9,11,13,15,17 respectively to obtain the next number.

2^nd term = 1^st term3 - 4; 3^rd term = 2^nd term3 - 4 and so on

Multiply the number by 2 and add 1 to it to get the next number.

Each number (except 100) can be expressed as 3 × power of 2: 12=3×4 (2^2), 48=3×16 (2^4), 384=3×128 (2^7), 768=3×256 (2^8), 3072=3×1024 (2^10). The number 100=3×33.33 doesn't fit this pattern and cannot be expressed as 3 times a power of 2. Alternatively, all others are divisible by 12 or 3, but 100 is not.

2 term = 1^st term1 + 1 ; 3^rd term = 2^nd term2 + 2  and so on

The numbers must be squares of (14, 13, 12, 11, 10, 9, 8. so 80 is wrong)

2^nd term = 1^st term1 + sqr of 1; 3^rd term = 2^nd term2 + sqr of 2 and so on

The pattern is +1 then ×(n/4) where n increases: 3(+1)→4(×2.25)→9(×2.5)→22.5(×3)→67.5(×3)→202.5(×4)→810. The number 3 breaks this pattern as it doesn't result from the same operation sequence - it's the starting seed. All other numbers follow the alternating +1 and multiply pattern.

2^nd term = 1^st term2 + 2 ; 3^rd term = 2^nd term2 + 4 ; 4^th term = 3^rd term*2 + 6 and so on

a * 2 + 1, b * 3 + 1, c*2 + 1, d * 3 + 1 and so on

2^nd term = 1^st term + sqr of 2 ; 3^rd term = 2^nd term + sqr of 3 and so on

Subtract 8 from the previous term and divide it by 2 to get the next term

2^nd term = 1st term2 - 2 ; 3^rd term = 2^nd term3 - 2 and so on

8^th term2 + 1 = 7^th term; 7^th term2 + 2 = 6^th term and so on

2^nd term = 1^st term1.5 ; 3^rd term = 2^nd term2 ; 4^th term = 3^rd term*2.5 and so on

2^nd term = ( 1^st term-30)/2 ; 3^rd term = (2^nd term-26)/2 and so on

2^nd term = 1^st term- sqr of 12 ; 3^rd term = 2^nd term - sqr of 10 and so on