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CHAPTER VI-XIXRD PROBLEMS.
169
Examples in illustration thereof. 293. The sum of the sories is 40; the number of terms is 5; and the common difference is 3; the first torm is not known now. (Find it out.) When the first term is 2, find out the cominon difference.
The rule for arriving at the aum and the number of terms in a series in arithinetical progression (with the aid of tho known lābha, which is the same as the quotiont obtainod by dividing the sum by the unknown number of terms therein) :
294. The libhre in diminished by the first term, and (thon) divided by the half of the common differonco; and on adding one to this samo resulting quantity), the number of terms in the serios (is obtained). The number of torins in the series multi plied by the läbha becomos the sum of the series.
An example in illustration thereof. 295. (There were a nuinber of utpalı flowers, representable a: the sum of a serios in arithmotical progression, whoroof) is the first term, and 3 the common difforence. A number of womon divided (these) utprla flowers (equally among them). Each woman had 8 for her share. How many were the women, and bow. many the flowers ?
The rule for arriving at the sum of the squares (of a given number of natural numbers beginning with one) :
296. The given number is increased by one, and (then) qnarod ; (this squared quantity is multiplied by two, and (then) diminished by the given quantity as inorcased by one. (The remainder thus
284. Algebraically,
+ 1, where l=
, which is the libha.
295. The number of women in this problem is conceivod to be equal to the number of terms in the series.
{ 2 (n+1)!-(n+1) 990. Algebraioally,
38, which is the sum of the squares of the natural numbers up to th.
