Question

find the sum of last 12 positive terms of 62,59,56,.............​

Answer 1

n/2(2a+(n-1)d=sum

12/2(2×62+11×-3)

6(124+-33)

6×91=546

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Answer 2

Answer:

The sum of last 12 positive terms = 222

Given problem:

find the sum of last 12 positive terms of 62,59,56,.............​

Step-by-step explanation:  

given number sequence  62, 59, 56 ...  which is in AP

                                                      [∵ common difference is equal ]

here first term a = 62  

common difference = T₂ - T₁ = 59 - 62 = -3

here to find sum of last 12 positive digits we need to find at which term the number value will be zero or will be a negative number  

let's assum that at nth term, the number value will be 0 or negative  

then  nth term of AP = a+(n-1)d $\leq$ 0

                              62 +(n-1)(-3) $\leq$ 0

                                 62 -3n +3 $\leq$ 0

                                     65 - 3n $\leq$ 0        

                                          65 $\leq$ 3n      

                                       21.666$\leq$ n          

for n > 21th term the number value will be negative    

⇒ in the given AP first 21 numbers are the positive terms

sum of last 12 positive terms

                 = (sum of first 21 terms) - (sum of first 9 terms)  

                 = $\frac{21}{2} [2(62)+(21-1)(-3) ] - \frac{9}{2} [ 2(62) +(9-1) (-3) ]$  

                 = 10.5 [ 124 +(20)(-3) ] - 4.5 [ 124 +(8)(-3) ]

                 = 10.5 [ 124 - 60 ] - 4.5 [124 - 24]  

                 =  10.5 (64) - 4.5 (100)

                 = 672 - 450 = 222