Question
In an A.P. 17th term is 7 more than its 10th term. Find the common difference.
✓
Answer
Given: t17 = 7 + t10 ……(1)
In t17, n = 17
In t10, n = 10
By using nth term of an A.P. formula,
tn = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
tn = nth term
Thus, on using formula in eq. (1) we get,
⇒ a + (17 – 1)d = 7 + (a + (10 – 1)d)
⇒ a + 16 d = 7 + (a + 9 d)
⇒ a + 16 d – a – 9 d = 7
⇒ 7 d = 7
$\Rightarrow d =\frac{7}{7}=1$
Thus, common difference “d” = 1
In t17, n = 17
In t10, n = 10
By using nth term of an A.P. formula,
tn = a + (n – 1)d
where n = no. of terms
a = first term
d = common difference
tn = nth term
Thus, on using formula in eq. (1) we get,
⇒ a + (17 – 1)d = 7 + (a + (10 – 1)d)
⇒ a + 16 d = 7 + (a + 9 d)
⇒ a + 16 d – a – 9 d = 7
⇒ 7 d = 7
$\Rightarrow d =\frac{7}{7}=1$
Thus, common difference “d” = 1
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the value of k for which the following systems of linear equations has an infinite number of solutions:
$2x + (k - 2)y = k,$
$6x + (2k - 1)y = (2k + 5).$
→$2x + (k - 2)y = k,$
$6x + (2k - 1)y = (2k + 5).$
Draw a circle of radius 3.6 cm. Draw a tangent to the circle at any point on it without using the centre.
Find the coordinates of centroid $G$ of $A B C$, if
(i) $A (8,9), B (4,5), C (6,2)$
(ii) $A (11,8), B (-6,5), C (1,-28)$
→(i) $A (8,9), B (4,5), C (6,2)$
(ii) $A (11,8), B (-6,5), C (1,-28)$
If $t_n=2 n-5$ is the nth term of an A.P., then find its first five terms.
→In three line segments OA, OB, and OC, points L, M, N respectively are so chosen that LM || AB and MN || BC but neither of L, M, N nor of A, B, C are collinear. Show that LN || AC.
In the given figure, the side of square is $28\ cm$ and radius of each circle is half of the length of the side of the square where O and O' are centres of the circles. Find the area of shaded region.
→Show that any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.
Solve the following quadratic equations by factorization:
$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3},$ $\text{x}\neq2, 4$
→$\frac{\text{x}-1}{\text{x}-2}+\frac{\text{x}-3}{\text{x}-4}=3\frac{1}{3},$ $\text{x}\neq2, 4$
In a $\triangle\text{ABC}$ right angled at B, $\angle\text{A}=\angle\text{C.}$ Find the values of.
$\sin\text{A}\sin \text{B}+\cos\text{A}\cos\text{B}$
→$\sin\text{A}\sin \text{B}+\cos\text{A}\cos\text{B}$
11, 8, 5, 2, . . . In this A.P. which term is number – 151?