Advertisements
Advertisements
Question
The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17. Find the progression.
Advertisements
Solution
\[\text { We have: } \]
\[ S_7 = 10\]
\[ \Rightarrow \frac{7}{2}\left[ 2a + (7 - 1)d \right] = 10\]
\[ \Rightarrow \frac{7}{2}\left[ 2a + 6d \right] = 10\]
\[ \Rightarrow a + 3d = \frac{10}{7} . . . (i)\]
\[\text { Also, the sum of the next seven terms } = S_{14} - S_7 = 17\]
\[ \Rightarrow \frac{14}{2}\left[ 2a + \left( 14 - 1 \right)d \right] - \frac{7}{2}\left[ 2a + (7 - 1)d \right] = 17\]
\[ \Rightarrow 7\left[ 2a + 13d \right]\]
\[ - \frac{7}{2}\left[ 2a + 6d \right] = 17\]
\[ \Rightarrow 14a + 91d - 7a - 21d = 17\]
\[ \Rightarrow 7a + 70d = 17\]
\[ \Rightarrow a + 10d = \frac{17}{7} . . . (ii)\]
\[\text { From (i) and (ii), we get }: \]
\[\frac{10}{7} - 3d = \frac{17}{7} - 10d\]
\[ \Rightarrow 7d = 1\]
\[ \Rightarrow d = \frac{1}{7}\]
\[\text { Putting the value in (i), we get: } \]
\[a + 3d = \frac{10}{7}\]
\[ \Rightarrow a + \frac{3}{7} = \frac{10}{7}\]
\[ \Rightarrow a = 1\]
\[ \therefore a = 1, d = \frac{1}{7}\]
The progression thus formed is
\[1, \frac{8}{7}, \frac{9}{7}, \frac{10}{7} . . .\]
APPEARS IN
RELATED QUESTIONS
In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.
Find the sum to n terms of the A.P., whose kth term is 5k + 1.
If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
if `a(1/b + 1/c), b(1/c+1/a), c(1/a+1/b)` are in A.P., prove that a, b, c are in A.P.
If the nth term an of a sequence is given by an = n2 − n + 1, write down its first five terms.
Let < an > be a sequence defined by a1 = 3 and, an = 3an − 1 + 2, for all n > 1
Find the first four terms of the sequence.
Let < an > be a sequence. Write the first five term in the following:
a1 = a2 = 2, an = an − 1 − 1, n > 2
Show that the following sequence is an A.P. Also find the common difference and write 3 more terms in case.
3, −1, −5, −9 ...
Show that the following sequence is an A.P. Also find the common difference and write 3 more terms in case.
9, 7, 5, 3, ...
Find:
18th term of the A.P.
\[\sqrt{2}, 3\sqrt{2}, 5\sqrt{2},\]
Find:
nth term of the A.P. 13, 8, 3, −2, ...
Is 68 a term of the A.P. 7, 10, 13, ...?
The first term of an A.P. is 5, the common difference is 3 and the last term is 80; find the number of terms.
If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term.
If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.
The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.
\[\text { If } \theta_1 , \theta_2 , \theta_3 , . . . , \theta_n \text { are in AP, whose common difference is d, then show that }\]
\[\sec \theta_1 \sec \theta_2 + \sec \theta_2 \sec \theta_3 + . . . + \sec \theta_{n - 1} \sec \theta_n = \frac{\tan \theta_n - \tan \theta_1}{\sin d} \left[ NCERT \hspace{0.167em} EXEMPLAR \right]\]
Find the sum of all even integers between 101 and 999.
Solve:
25 + 22 + 19 + 16 + ... + x = 115
Find the r th term of an A.P., the sum of whose first n terms is 3n2 + 2n.
Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.
The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.
If a, b, c is in A.P., then show that:
a2 (b + c), b2 (c + a), c2 (a + b) are also in A.P.
The income of a person is Rs 300,000 in the first year and he receives an increase of Rs 10000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.
Write the common difference of an A.P. whose nth term is xn + y.
Write the common difference of an A.P. the sum of whose first n terms is
If Sn denotes the sum of first n terms of an A.P. < an > such that
If the sum of n terms of an A.P. is 2 n2 + 5 n, then its nth term is
Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn − k Sn − 1 + Sn − 2 , then k =
Mark the correct alternative in the following question:
The 10th common term between the A.P.s 3, 7, 11, 15, ... and 1, 6, 11, 16, ... is
If second, third and sixth terms of an A.P. are consecutive terms of a G.P., write the common ratio of the G.P.
Write the quadratic equation the arithmetic and geometric means of whose roots are Aand G respectively.
The first term of an A.P. is a, the second term is b and the last term is c. Show that the sum of the A.P. is `((b + c - 2a)(c + a))/(2(b - a))`.
If the sum of n terms of an A.P. is given by Sn = 3n + 2n2, then the common difference of the A.P. is ______.
If in an A.P., Sn = qn2 and Sm = qm2, where Sr denotes the sum of r terms of the A.P., then Sq equals ______.
If n AM's are inserted between 1 and 31 and ratio of 7th and (n – 1)th A.M. is 5:9, then n equals ______.
