• # question_answer In which of the following situations, does the list of numbers involved does not make an arithmetic progression? (i) The taxi fare after each km when the fare is Rs. 20 for the first km and Rs. 11 for each additional km. (ii) The amount of air present in a cylinder when a vacuum pump removes $\frac{1}{4}$of the air remaining in the cylinder at a time. (iii) The cost of digging a well after every metre of digging, when it cost Rs. 250 for the metre and rises by Rs. 40 for each subsequent metre. (iv) The amount of money in the account every year, when Rs. 8000 is deposited at compound interest at 10% per annum. A)  (i) & (ii)                       B)  (ii) & (iv)   C)  (iii) & (i)                       D)  (i) & (iv)

(b): (i) According to question, the fare for journey of first km i.e., 1 km is Rs. 20 and next 2 km, 3 km, 4 km...are respectively Rs.$\left( 20+11 \right)$, Rs.$\left( 20+2\times 11 \right)$, Rs.$\left( 20+3\times 11 \right)$, ......so on. i.e. 20, 31, 42, 53 and so on Here, each term is obtained by adding 11 to the preceding term except first term. So, it is an AP. (ii)  Let the amount of air present in the cylinder by y units. The air present in the cylinder is given by $y,y-\frac{y}{4}=\frac{3y}{4},\frac{3y}{4}-\frac{1}{4}\times \frac{3y}{4}=\frac{12y-3y}{16}=\frac{9y}{16},......$ Or, $y,\frac{3y}{4},\frac{9y}{16},.....$ Here, $\frac{3y}{4}-y=-\frac{y}{4}$                  $\left( \because {{T}_{2}}-{{T}_{1}} \right)$ And $\frac{9y}{16}-\frac{3y}{4}=\frac{9y-12y}{16}=-\frac{3y}{16}$       $\left( \because {{T}_{3}}-{{T}_{2}} \right)$ $\Rightarrow {{T}_{2}}-{{T}_{1}}\ne {{T}_{3}}-{{T}_{2}}$ $\Rightarrow$It does not form an AP, because common difference is not the same. If you look it carefully, it forms a G.P. (not an A.P.) (iii) The cost of digging for the first metre, second metre, third metre and so on are respectively Rs. 250 (Rs. 250 + 40). Rs (290 + 40)......and so on. Here, each term is obtained by adding 40 to the preceding term, So, it is an AP. (iv) According to question, the amount of money in the account in the first year, second year, third year and so on are respectively. $8000,8000\left( 1+\frac{8}{100} \right),8000{{\left( 1+\frac{8}{100} \right)}^{2}}$,???. Clearly, it is a G.P. (not an A.P.)