Show that a1 , a2 , . . ., an, . . . form an AP where an is def

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 Multiple Choice QuestionsShort Answer Type

241.

Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

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242.

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

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243.

Show that a1 , a2 , . . ., an, . . . form an AP where an is defined as below :
a= 3 + 4n


It is given that,

         a= 3 + 4n
So,    a= 3 + 4(1) = 3 + 4 = 7
         a2  = 3 + 4 (2) = 3 + 8 = 11
and    a= 3 + 4(3) = 3 + 12 = 15
Now, er have following numbers :
    7, 11, 15, ...........
Since the difference between each pair of consecutive terms are constant

So, the given number forms an A. P.
Here, a = 7, d = 4,    n  = 15
We know that, 

straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket
So comma space space space straight S subscript 25 equals 25 over 2 left square bracket 2 space straight x space 7 space plus space left parenthesis 25 space minus space 1 right parenthesis space 4 right square bracket
space space space space space space space space space space space space space space equals 25 over 2 left square bracket 14 plus 96 right square bracket space equals space 25 over 2 cross times 110
space space space space space space space space space space space space space space space equals space 25 cross times 55 space equals space 1375

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244.

Show that a1 , a2 , . . ., an, . . . form an AP where an is defined as below :
an = 9 – 5n

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245.

If the sum of the first n terms of an AP is 4n – n2 , what is the first term (that is S1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

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246.

Find the sum of the first 40 positive integers divisible by 6.

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247.

Find the sum of the first 15 multiples of 8.

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248.

Find the sum of the odd numbers between 0 and 50.

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249. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows : Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much does a delay of 30 days cost the contractor?
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250. A sum of Rs. 280 is to be used to award four prizes. If each prize after the first is Rs. 20 less than the next most valuable one, find the value of each of the prizes.
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